Motion under Gravity is part of Physics. Below are notes on 11 sub-topics, followed by practice MCQs from State PSC and UPSC papers.

Concept of air resistance and its influence on motion under gravity

  • Ideal Motion under Gravity (Vacuum/ Absence of Air Resistance):
    • Only gravitational force acts on the object.
    • Acceleration due to gravity ('g') is constant and independent of the object's mass, size, or shape.
    • All objects, regardless of their mass or material, fall at the same rate and take the same time to reach the ground when dropped from the same height.
    • Example: Objects falling on the Moon (no atmosphere) or in a vacuum chamber.
  • Air Resistance:
    • A form of fluid friction exerted by air on an object moving through it, always opposing the direction of motion.
    • Its magnitude depends on the object's speed (increases with speed), shape, frontal area, and the density of the air.
  • Motion under Gravity with Air Resistance (Earth's Atmosphere):
    • An object's motion is influenced by two forces: gravitational force (downwards) and air resistance (upwards).
    • The net force on the object is (Gravitational Force - Air Resistance).
    • Consequently, objects do not fall at the same rate; objects with greater massto-frontal-area ratio (e.g., a rock) fall faster than objects with a smaller ratio (e.g., a feather or a sheet of paper).
    • Terminal Velocity: As an object falls, its speed increases, and thus air resistance increases. When the upward air resistance equals the downward gravitational force, the net force becomes zero. The object then falls at a constant maximum speed called terminal velocity (zero acceleration).

Conservation of mechanical energy in motion under gravity

  • Total Mechanical Energy (Emech) is the sum of Potential Energy (PE) and Kinetic Energy (KE). Emech = PE + KE.
  • Potential Energy (PE) = mgh, where m = mass, g = acceleration due to gravity, and h = height above a reference level.
  • Kinetic Energy (KE) = (1/2)mv², where m = mass and v = velocity.
  • In motion under gravity (free fall), neglecting air resistance, the total mechanical energy (PE + KE) of a body remains constant. This is the Principle of Conservation of Mechanical Energy.
  • During free fall:
    • Potential Energy (PE) decreases as the height of the body decreases.
    • Kinetic Energy (KE) increases as the speed of the body increases.
    • The decrease in PE is equal to the increase in KE, ensuring the total mechanical energy is conserved.
  • If a body is dropped from rest at a height hinitial:
    • Its initial KE is 0.
    • Its total mechanical energy is equal to its initial PE: Emech = mghinitial.
  • At any intermediate height h during free fall, Emech = mgh + (1/2)mv². Since Emech is constant, mghinitial = mgh + (1/2)mv².
  • When the body reaches the ground (h=0):
    • Its PE becomes 0.
    • Its total mechanical energy is entirely converted into KE: Emech = (1/2)mvfinal².
    • Therefore, mghinitial = (1/2)mvfinal², which can be used to find the final speed (vfinal = √(2ghinitial)).
  • Units: Energy in Joules (J), mass in kilograms (kg), height in meters (m), velocity in meters/second (m/s), and acceleration due to gravity in meters/second² (m/s²).

Effect of Earth's rotation and latitude on the value of 'g'

  • The acceleration due to gravity ('g') varies due to Earth's shape, altitude, depth, and rotation.
  • Effect of Earth's Shape: Earth is an oblate spheroid (equatorial radius > polar radius). Since `g = GM/R²`, 'g' is greater at poles than at the equator (smaller R at poles).
  • Effect of Earth's Rotation:
    • Earth's rotation creates a centrifugal force, reducing the effective 'g'.
    • Effective 'g' at latitude λ is `g' = g ω²Rcos²λ`, where ω is angular velocity and R is Earth's radius.
    • At Poles (λ=90°): `cos²90° = 0`, so `g' = g`. Rotation has no effect; 'g' is maximum.
    • At Equator (λ=0°): `cos²0° = 1`, so `g' = g - ω²R`. 'g' is minimum.
    • 'g' increases as latitude increases from equator to poles.
  • Combined Effect: 'g' is maximum at poles and minimum at the equator.
  • Pendulum Clock Time Period: For a simple pendulum, `T = 2π√(L/g)`.
  • This implies `T ∝ 1/√g`. If 'g' decreases, 'T' increases (clock runs slower). If 'g' increases, 'T' decreases (clock runs faster).
  • On the Moon, `gmoon ≈ gearth / 6`.
  • Therefore, `Tmoon / Tearth = √(gearth / gmoon) = √(gearth / (gearth/6)) = √6`.
  • Since `Tmoon = √6 Tearth`, the pendulum on the Moon takes longer to complete one oscillation. Thus, the clock will run `√6` times slower.

Equations of motion for objects under free fall

  • Free Fall: Motion of an object under the exclusive influence of gravity, with air resistance considered negligible.
  • Acceleration due to Gravity (g): In free fall, objects experience a constant acceleration 'g', approximately 9.8 m/s² (downwards). This makes free fall an example of uniformly accelerated motion.
  • Vacuum Effect: In a vacuum (absence of air resistance), all objects, irrespective of their mass, size, or shape, fall with the same acceleration 'g' and reach the ground simultaneously if released from the same height.
  • Equations of Motion (for constant acceleration 'g'):
    • Downward Motion (taking downward as positive):
      • Final Velocity: `v = u + gt`
      • Displacement/Height: `h = ut + (1/2)gt^2`
      • Velocity-Displacement Relation: `v^2 = u^2 + 2gh`
    • Upward Motion (taking upward as positive, 'g' acts downwards so it's negative):
      • Final Velocity: `v = u - gt`
      • Displacement/Height: `h = ut (1/2)gt^2`
      • Velocity-Displacement Relation: `v^2 = u^2 - 2gh`
    • Where `u` = initial velocity, `v` = final velocity, `h` = displacement/height, `t` = time.
  • Object Dropped from Rest: If an object is dropped, its initial velocity `u = 0`.
    • `v = gt`
    • `h = (1/2)gt^2`
    • `v^2 = 2gh`
  • Force in Free Fall: The net force acting on an object of mass `m` in free fall is its weight, `F = mg`, directed downwards. If upward direction is taken as positive, `F = -mg`.
  • Weightlessness: Occurs during free fall (e.g., in a lift whose cable breaks). The apparent weight of an object becomes zero because its acceleration `a` equals `g` (i.e., `Wapparent = m(g - a) = m(g - g) = 0`).
  • Motion in an Elevator with Uniform Velocity: If an elevator moves with constant velocity (zero acceleration), the acceleration experienced by an object dropped inside it is still `g`. Consequently, the time taken for the object to fall a certain height relative to the elevator floor is identical to when the elevator is stationary.

Gravitational acceleration (g) and its assumed constancy near Earth's surface

  • Gravitational acceleration (g) is the acceleration experienced by an object due to the gravitational pull of a celestial body.
  • Near Earth's surface, the approximate value of g is 9.8 m/s². For practical purposes in problems, it is often taken as 10 m/s².
  • The acceleration due to gravity (g) is assumed constant for motions occurring over small heights near the Earth's surface.
  • The formula for 'g' on a celestial body is given by `g = GM/R²`, where G is the universal gravitational constant, M is the mass of the body, and R is its radius. Thus, 'g' depends on both the mass and radius of the celestial body.
  • Free Fall: Refers to the motion of an object under the sole influence of gravity, with negligible air resistance.
  • In free fall (or a vacuum), all objects, regardless of their mass, fall with the same acceleration (g) and reach the ground simultaneously.
  • When an object is dropped freely, its initial velocity (u) is zero.
  • When an object is thrown vertically upwards, its final velocity (v) becomes zero at its maximum height.
  • Gravity acts continuously on an object, causing it to accelerate downwards. This causes objects thrown upwards to eventually fall back down.
  • During free fall, while acceleration (g) remains constant (near the surface), velocity, momentum, kinetic energy, and potential energy are not uniform and continuously change.
  • Escape Velocity (Vₑ): The minimum velocity an object needs to completely escape the gravitational pull of a celestial body without further propulsion.
  • The formula for escape velocity is `Vₑ = √(2gR)`, where 'g' is acceleration due to gravity and 'R' is the radius of the celestial body.
  • For Earth, the escape velocity is approximately 11.2 km/s.

Maximum height and time of flight for vertically projected objects

  • For vertically projected objects: Acceleration due to gravity (`g`) acts downwards. During upward motion, effective acceleration `a = g`.
  • At the maximum height, the final vertical velocity (`v`) of the object is zero.
  • Time to reach maximum height (`th`): Given initial upward velocity (`u`), `th = u/g`.
  • Maximum height (`H`): Given initial upward velocity (`u`), `H = u² / (2g)`.
  • Maximum height (`H`): Given time to reach maximum height (`th`), `H = (1/2)gth²`.
  • Time of Flight (`T`): Total time taken from projection to return to the initial projection level. `T = 2 th = 2u/g`.
  • Value of `g`: Standard value is approximately `9.8 m/s²`. For simplicity in problems, it is often taken as `10 m/s²`.

Projectile motion: horizontal projection from a height

  • Object thrown horizontally from a height 'h' with initial horizontal velocity 'u'.
  • Initial vertical velocity (uy) = 0.
  • Horizontal Motion:
    • Horizontal velocity (Vx) remains constant: Vx = u (neglecting air resistance).
    • Horizontal displacement (Range, R) = u × t (where 't' is time of flight).
  • Vertical Motion:
    • Vertical acceleration (ay) = g (downwards).
    • Vertical displacement (h) = (1/2)gt² (since uy = 0).
    • Vertical velocity (Vy) at time 't' = gt.
  • Time of Flight (t):
    • Calculated from vertical motion: t = √(2h/g).
    • Crucially, time of flight is independent of the initial horizontal velocity and mass of the projectile.
    • Therefore, objects projected horizontally from the same height, irrespective of their horizontal speeds or masses, will hit the ground simultaneously.
  • Horizontal Range (R): R = u × √(2h/g).
  • Velocity at any instant 't':
    • Magnitude: V = √(u² + (gt)²).
    • Direction: tan θ = Vy / Vx = gt / u (angle with horizontal).
  • Path of projectile: Parabolic.
  • Relative Motion: An observer (e.g., pilot) moving with the same initial horizontal velocity as the projectile (e.g., a dropped bomb) will observe the object falling vertically downward.

Projectile motion: projection at an angle to the horizontal

  • Definition: Motion of an object (projectile) projected at an angle to the horizontal, acted upon only by gravity (air resistance neglected).
  • Independence of Motion: Horizontal and vertical components of motion are independent.
  • Horizontal Motion: Velocity component (vx = u cosθ) is constant; acceleration (ax) = 0.
  • Vertical Motion: Initial velocity component (vy = u sinθ); constant downward acceleration (ay = -g).
  • Vertical Drop due to Gravity: For any projectile, the vertical distance it falls from its hypothetical straight-line path (no gravity) in time 't' is (1/2)gt2.
  • Trajectory: Parabolic path.
  • Time of Flight (T): T = (2u sinθ)/g.
  • Maximum Height (H): H = (u2 sin2θ)/(2g).
  • Horizontal Range (R): R = (u2 sin(2θ))/g. Maximum range (Rmax = u2/g) occurs at θ = 45°. Same range for complementary angles (θ and 90°-θ).
  • Crucial Principle (for "Monkey and Hunter" problem): If a projectile is aimed directly at a target, and the target simultaneously begins free-fall (drops) from its position, the projectile will hit the target. This occurs because both the projectile and the target fall by the same vertical distance (1/2)gt2 in the time it takes for the projectile to reach the target's original horizontal position.

Range, maximum height, and time of flight for angled projectile motion

  • Projectile Motion: Motion of an object projected into the air, subject only to constant gravitational acceleration `g` (neglecting air resistance).
  • Initial Conditions: Object projected with initial velocity `u` at an angle `θ` with the horizontal.
  • Velocity Components:
    • Horizontal velocity `ux = u cosθ` (constant).
    • Vertical velocity `uy = u sinθ`.
  • Acceleration: Horizontal acceleration `ax = 0`; Vertical acceleration `ay = -g` (downwards).
  • Time of Flight (T): Total time the projectile remains in the air until it returns to the same horizontal level.
    • Formula: `T = (2u sinθ) / g`
    • Time to reach maximum height `th = T/ 2 = (u sinθ) / g`.
  • Maximum Height (H): Highest vertical distance reached from the point of projection.
    • Formula: `H = (u^2 sin^2θ) / (2g)`
    • `H` increases as the angle `θ` increases from 0° to 90°.
  • Horizontal Range (R): Total horizontal distance covered by the projectile until it returns to the same horizontal level.
    • Formula: `R = (u^2 sin(2θ)) / g`
    • Maximum Range `Rmax` occurs at `θ = 45°` for a given `u`. `Rmax = u^2/g`.
    • Projectiles thrown at complementary angles (e.g., `θ` and `90° - θ`) will have the same range `R` if their initial speeds `u` are identical.
    • For complementary angles, the projectile with the larger angle will have a greater maximum height and a longer time of flight.

Variation of 'g' with altitude and depth

  • Acceleration due to gravity ('g') is maximum at the Earth's surface.
  • Acceleration due to gravity decreases with increasing altitude (height above the Earth's surface).
  • Acceleration due to gravity decreases with increasing depth (below the Earth's surface).
  • At the center of the Earth (maximum depth), the acceleration due to gravity becomes zero.
  • The weight of a body (W=mg) decreases with both increasing altitude and increasing depth, as 'g' decreases, while its mass remains constant.

Vertical motion: upward and downward projection of objects

  • Motion under gravity means constant acceleration `g` (approx. 9.8 m/s² or 10 m/s²), always directed downwards.
  • For vertical motion, the standard equations of motion are adapted, often using sign conventions:
    • If upward direction is taken as positive: `v = u - gt`, `h = ut - (1/2)gt²`, `v² = u² - 2gh`.
    • If downward direction is taken as positive (e.g., for free fall or downward projection): `v = u + gt`, `h = ut + (1/2)gt²`, `v² = u² + 2gh`.
  • For an object projected vertically upward with initial speed `u`:
    • Velocity at the maximum height is `0`.
    • Time to reach maximum height (`tasc`) = `u/g`.
    • Maximum height (`Hmax`) = `u²/(2g)`.
    • Total time of flight (to return to the projection point) = `2u/g`.
    • Speed upon returning to the point of projection is equal to the initial projection speed `u`.
  • When two objects collide, their positions relative to a common reference point are equal at the time of collision (`s1(t) = s2(t)`).
  • For bouncing objects:
    • The speed of the object immediately before impact (`vimpact`) is calculated using motion equations.
    • If a percentage of speed is lost upon impact, the speed after bounce (`vbounce`) = `vimpact × (1 - fractional speed loss)`.
    • The height reached after bouncing (`hbounce`) can be calculated using `hbounce = vbounce² / (2g)`.

5. Physical Properties of Materials

Practice MCQs: Motion under Gravity

Concept of air resistance and its influence on motion under gravity

  1. Which one of the following statements is not correct for a freely falling object?

    Asked in: UPSC Geoscientist 2022

    1. It accelerates.
    2. Its momentum keeps on changing.
    3. Its motion is affected only by the gravity.
    4. Its motion is affected both by the gravity as well as by the air resistance.
    Show answer

    Answer: D. Its motion is affected both by the gravity as well as by the air resistance.

    - Free fall is motion under the sole influence of gravity. - Air resistance is typically ignored in the definition of free fall. - A freely falling object accelerates and its momentum changes.

  2. An iron ball and a wooden ball of the same radius are released from a height 'H' in vacuum. The time taken to reach the ground will be-

    Asked in: ASCO 2014

    1. more for the iron ball
    2. more for the wooden ball
    3. equal for both
    4. in the ratio of their weights
    Show answer

    Answer: C. equal for both

    HORTICULTURE DEVELOPMENT OF‐ FICER(HDO) -2015 In a vacuum, the acceleration due to gravity is independent of the mass and composition of the object.

  3. An astronaut drops a hammer and a feather on the moon. What will happen?

    Asked in: Paper I 6

    1. Both will fall at the same rate
    2. The hammer will hit the ground first, but not as fast as on Earth
    3. The hammer will fall faster because it is heav‐ ier
    4. The feather will float, and the hammer will fall
    Show answer

    Answer: A. Both will fall at the same rate

    In a vacuum (like the Moon's surface), all objects, regardless of their mass or shape, fall with the same acceleration due to gravity. Air resistance is the only factor that causes lighter objects to fall slower on Earth.

Conservation of mechanical energy in motion under gravity

  1. A stone of mass 1 kg and initially at rest is dropped from a tower of height 40 m. When it reaches a height of 10 m from the ground level, what will be the values of its potential energy (PE) and kinetic energy (KE)? (Acceleration due to gravity is 10 m/s²)

    Asked in: UPSC Geoscientist 2021

    1. PE = 300 J, KE = 100 J
    2. PE = 200 J, KE = 200 J
    3. PE = 100 J, KE = 300 J
    4. PE = 100 J, KE = 200 J
    Show answer

    Answer: C. PE = 100 J, KE = 300 J

    In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.

  2. A body has a free fall from a height of 20 m. After falling through a distance of 5 m, the body would

    Asked in: UPSC NDA 1 2016

    1. lose one-fourth of its total energy
    2. lose one-fourth of its potential energy
    3. gain one-fourth of its potential energy
    4. gain three-fourth of its total energy
    Show answer

    Answer: B. lose one-fourth of its potential energy

    In free fall (assuming no air resistance), mechanical energy is conserved. As potential energy decreases, kinetic energy increases by an equal amount. Potential energy is proportional to height.

  3. A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9.8 m/s²)

    Asked in: UPSC NDA 2 2019

    1. 980 m.s⁻¹ and 980 J
    2. √980 m.s⁻¹ and √980 J
    3. 980 m.s⁻¹ and √980 J
    4. √980 m.s⁻¹ and 980 J
    Show answer

    Answer: D. √980 m.s⁻¹ and 980 J

    The problem involves calculating the final speed of a falling object using kinematics and determining the total mechanical energy at the beginning of the fall using the principle of conservation of energy.

  4. There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is

    Asked in: UPSC NDA 2 2024

    1. 62·5 m/s
    2. 2·0 m/s
    3. 50 m/s
    4. 40 m/s
    Show answer

    Answer: A. 62·5 m/s

    When the ball is released freely from a height, its potential energy is converted into kinetic energy as it falls. By the principle of conservation of mechanical energy (assuming no air resistance), the potential energy at the initial height is equal to the kinetic energy just before hitting the ground.

  5. When a body falls freely towards the earth, then its total energy :

    Asked in: OPSC CS Prelims 2019

    1. Increases
    2. Remains same
    3. Decreases
    4. First increases and then decreases
    Show answer

    Answer: B. Remains same

    - Free fall is motion under the influence of gravity alone. - Total mechanical energy (E) = Kinetic Energy (KE) + Potential Energy (PE). - As the body falls, PE (mgh) decreases because h decreases, and KE (1/2 mv²) increases because speed v increases. - In the absence of non-conservative forces like air resistance, the decrease in PE is exactly equal to the increase in KE, keeping the total energy constant.

  6. During free fall of an object :

    Asked in: UPSC CAPF 2010

    1. its potential energy increases and its kinetic energy decreases.
    2. its potential energy decreases and its kinetic energy increases.
    3. both its potential energy and kinetic energy in‐ crease.
    4. both its potential energy and kinetic energy de‐ crease.
    Show answer

    Answer: B. its potential energy decreases and its kinetic energy increases.

    - Free fall is defined as motion under the influence of gravity only. - Potential energy (PE) due to gravity is given by mgh, where m is mass, g is acceleration due to gravity, and h is height. As an object falls, its height (h) decreases, so its potential energy decreases. - Kinetic energy (KE) is given by 0.5mv², where m is mass and v is velocity. As an object falls under gravity, its speed (v) increases (due to acceleration 'g'), so its kinetic energy increases. - In ideal free fall (neglecting air resistance), the total mechanical energy (PE + KE) is conserved. The decrease in potential energy is equal to the increase in kinetic energy.

  7. When a body of certain mass falls freely from a height it's- Potential energy decreases and Kinetic energy increases. Potential energy increases and Kinetic energy decreases. Potential energy and Kinetic energy both increase. Potential energy and Kinetic energy both decrease.

    Asked in: School Lecturer (School Edu.) 2018 1 2018

    1. (1)
    2. (2)
    3. (3)
    4. (4)
    Show answer

    Answer: A. (1)

    During free fall, potential energy decreases as height decreases, and kinetic energy increases as velocity increases, due to the conversion of potential energy to kinetic energy.

Effect of Earth's rotation and latitude on the value of 'g'

  1. A pendulum clock that keeps correct time on the earth, is taken to moon. It will then run

    Asked in: WBPSC CS Prelims 2021

    1. at correct rate.
    2. 6 times faster.
    3. √6 times faster.
    4. √6 times slower.
    Show answer

    Answer: D. √6 times slower.

    - T_earth = 2π√(L/g_earth) - T_moon = 2π√(L/ g_moon) = 2π√(L / (g_earth/6)) = 2π√(6L/g_earth) = √6 (2π√(L/g_earth)) = √6 T_earth - The period on the Moon (T_moon) is √6 times longer than on Earth (T_earth). - A pendulum clock measures time by counting oscillations. A longer period means each swing takes longer, so the clock will run slower.

Equations of motion for objects under free fall

  1. If a feather, a rubber ball and a wooden ball are falling freely simultaneously from the same height in vacuum, then

    Asked in: BPSC Prelims Exam 67th Cancelled

    1. the feather will reach at the ground first
    2. the rubber ball will reach at the ground first
    3. the wooden ball will reach at the ground first
    4. all the three will reach at the ground together
    Show answer

    Answer: D. all the three will reach at the ground together

    In a vacuum, there is no air resistance. The only force acting on a falling object is gravity. According to Galileo's principle of falling bodies, objects in free fall under gravity accelerate at the same rate regardless of their mass or composition, provided air resistance is negligible or absent (as in a vacuum). The acceleration due to gravity (g) is constant for all objects at a given location.

  2. A man weighing 70 kg is coming down in a lift. If the cable of the lift breaks suddenly, the weight of the man would become

    Asked in: UPSC NDA 1 2016

    1. 70 kg
    2. 35 kg
    3. 140 kg
    4. Zero
    Show answer

    Answer: D. Zero

    Weight is the force of gravity acting on an object's mass (W = mg). Apparent weight is the normal force exerted by the supporting surface on the object. When the lift cable breaks, the lift and everything inside it are in free fall, accelerating downwards at the acceleration due to gravity (g). In a state of free fall, there is no supporting force counteracting gravity. Therefore, the apparent weight of the man (the force exerted by the lift floor on him) becomes zero. This is the sensation of weightlessness. His actual weight (mass g) remains the same, but his apparent weight is zero.

  3. In a vacuum, a five-rupee coin, a feather of a sparrow bird and a mango are dropped simultaneously from the same height. The time taken by them to reach the bottom is t1, t2 and t3 respectively. In this situation, we will observe that

    Asked in: UPSC NDA 2 2017

    1. t1 > t2 > t3
    2. t1 > t3 > t2
    3. t3 > t1 > t2
    4. t1 = t2 = t3
    Show answer

    Answer: D. t1 = t2 = t3

    - In a vacuum, acceleration due to gravity is constant for all objects ($g$). - Absence of air resistance. - Time taken to fall from the same height is independent of mass.

  4. The motion of a particle of mass m is described by the relation, $y = ut - \frac{1}{2} gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is

    Asked in: UPSC NDA 2 2023

    1. $F = m\left(\frac{du}{dt}\right)$
    2. $F = mg$
    3. $F = m\left(\frac{dy}{dt}\right)$
    4. $F = -mg$
    Show answer

    Answer: D. $F = -mg$

    The given equation of motion, $y = ut - \frac{1}{2} gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\frac{dy}{dt} = u - gt$. The second derivative gives acceleration: $\frac{d^2y}{dt^2} = -g$. According to Newton's second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.

  5. Which one of the following holds true for a freely falling object ?

    Asked in: UPSC CAPF 2023

    1. It moves with a uniform velocity.
    2. It moves with a uniform speed.
    3. It moves with a non-uniform acceleration.
    4. It moves with a uniform acceleration.
    Show answer

    Answer: D. It moves with a uniform acceleration.

    - Free fall is defined as motion under the sole influence of gravity. - Near the Earth's surface, the acceleration due to gravity ($\text{g}$) is approximately constant (about 9.8 m/s²). - This constant acceleration causes the velocity of the object to change uniformly over time.

  6. A ball is dropped from the top of a high building with a constant acceleration of 9.8 m/ s2. What will be its velocity after 3 seconds?

    Asked in: UPPCS Prelims 1998

    1. 9.8 m/s
    2. 19.6 m/s
    3. 29.4 m/s
    4. 39.2 m/s
    Show answer

    Answer: C. 29.4 m/s

    This is a problem involving uniformly accelerated motion. Given: Initial velocity (v₀) = 0 m/s (since the ball is dropped) Acceleration (a) = 9.8 m/s² (acceleration due to gravity, assumed constant) Time (t) = 3 seconds The formula relating final velocity (v), initial velocity (v₀), acceleration (a), and time (t) is v = v₀ + at. Substituting the given

  7. A person standing on the floor of an elevator drop a coin. The coin reaches the floor of the elevator in a time t1 when the elevator is stationary and in time t2 when it is moving with uniform speed, then - (1) The relation between t1 and t2 depends on whether lift is moving upward or downward. (2) t1 > t2 (3) t1 < t2 (4) t1 = t2 (5) Question not attempted

    Asked in: Exam 2024 Group a 2024

    1. (1) is correct
    2. (2) is correct
    3. (3) is correct
    4. (4) is correct
    Show answer

    Answer: D. (4) is correct

    Lecturer and Coach (School Edu. Dept.) Comp. When an object is dropped inside an elevator, its motion is governed by gravity relative to the elevator. If the elevator is stationary or moving with a uniform speed (either upwards or downwards), the acceleration due to gravity acting on the coin is the same. Since the initial conditions (dropped from rest relative to the elevator) and the acceleration due to gravity are the same in both scenarios, the time taken for the coin to reach the floor will be the same. This is because the physics inside the uniformly accelerating (or non-accelerating) frame of reference is the same as in an inertial frame of reference (Galilean relativity).

  8. An object covers equal distances in 2 s, 4 s and 8 s in successive intervals of time. What is its motion?

    Asked in: lower devision clerk 2025

    1. Uniform motion
    2. Uniformly accelerated
    3. Retarded
    4. Simple harmonic
    Show answer

    Answer: B. Uniformly accelerated

    The observation that an object takes longer to cover successive equal distances indicates a decreasing speed. This is characteristic of uniformly accelerated motion where the acceleration is negative (uniform retardation). While the specific numbers in the question lead to a mathematical inconsistency for non-zero acceleration, the fundamental nature of the motion described points towards uniform acceleration.

  9. An iron ball and a wooden ball of the same radius are released from a height 'H' in vacuum. The time taken to reach the ground will be–

    Asked in: POST GRADUATE TEACHER (PGT) 2018

    1. more for the iron ball
    2. more for the wooden ball
    3. equal for both
    4. in the ratio of their weights
    Show answer

    Answer: C. equal for both

    In a vacuum, all objects fall at the same rate due to gravity, irrespective of their mass or density, meaning they will reach the ground simultaneously if released from the same height.

Gravitational acceleration (g) and its assumed constancy near Earth's surface

  1. Two objects of different masses falling freely near the surface of the Moon would

    Asked in: BPSC Prelims Exam 69th

    1. have different accelerations
    2. undergo a change in their inertia
    3. have same velocity at any instant
    4. experience forces of same magnitude
    Show answer

    Answer: C. have same velocity at any instant

    - The acceleration due to gravity is independent of the mass of the falling object in free fall. - On the Moon (vacuum), all objects fall with the same constant acceleration. - Objects starting from rest with the same constant acceleration will have the same velocity at the same elapsed time.

  2. An object is dropped from a height onto the floor. Which one of the following remains uniform as it falls ?

    Asked in: UPSC CDS 2 2022

    1. Its acceleration
    2. Its momentum
    3. Its kinetic energy
    4. Its potential energy
    Show answer

    Answer: A. Its acceleration

    Acceleration due to gravity (g) is approximately 9.8 m/s² near the Earth's surface and acts downwards. For a freely falling object (with negligible air resistance), this acceleration is constant.

  3. The acceleration due to gravity at the Earth's surface depends on

    Asked in: UPSC CDS 2 2022

    1. its mass only.
    2. its radius only.
    3. both its mass and radius.
    4. either its mass or its radius.
    Show answer

    Answer: C. both its mass and radius.

    The acceleration due to gravity (g) at the surface of a celestial body like Earth is given by Newton's Law of Universal Gravitation. The formula for g is g = GM/R², where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth. This formula clearly shows that the acceleration due to gravity depends directly on the mass (M) and inversely on the square of the radius (R).

  4. Which of the following statement(s) about the freely falling objects is/are FALSE? Freely falling objects are the objects falling from a height towards the earth under gravitational force of the earth. The acceleration of a body falling freely towards the earth depends on the mass of the body. If an object is dropped freely from a height, its initial velocity 'u' is zero and if an object is thrown vertically upwards, its final velocity 'v' becomes zero. Choose the correct answer from the options given below:

    Asked in: HAS Prelims 2022

    1. 1 and 3 only
    2. 2 only
    3. 3 only
    4. 2 and 3 only
    Show answer

    Answer: B. 2 only

    In a vacuum, all objects fall with the same acceleration due to gravity, regardless of their mass. The acceleration due to gravity (g) is approximately 9.8 m/s² on Earth.

  5. If g is the acceleration due to gravity and R is the radius of the earth, then escape velocity is given by the relation

    Asked in: TNPSC CS Prelims 2015

    1. $V_e = \sqrt{2g}$
    2. $V_e = 2\sqrt{g}$
    3. $V_e = \sqrt{2Rg}$
    4. $V_e = 2\sqrt{Rg}$
    Show answer

    Answer: C. $V_e = \sqrt{2Rg}$

    The potential energy of an object of mass m at a distance R from the center of a body of mass M is $U = -\frac{GMm}{R}$. The kinetic energy is $K = \frac{1}{2}mV_e^2$. For the object to escape, its total mechanical energy must be at least zero at infinity, i.e., $\frac{1}{2}mV_e^2 - \frac{GMm} {R} \ge 0$. The minimum velocity occurs when the equality holds: $\frac{1}{2}mV_e^2 = \frac{GMm}{R}$. This simplifies to $V_e^2 = \frac{2GM}{R}$. Since $GM = gR^2$, substituting this gives $V_e^2 = \frac{2(gR^2)}{R} = 2gR$. Therefore, $V_e = \sqrt{2gR}$.

  6. In vaccum all objects fall with

    Asked in: TNPSC CS Prelims 2019

    1. same acceleration and heavy bodies reach ground first
    2. same acceleration and reach the ground at the same time
    3. different acceleration and reach the ground at different time
    4. different acceleration and heavy bodies reach ground first
    Show answer

    Answer: B. same acceleration and reach the ground at the same time

    In a vacuum, where there is no air resistance, all objects fall under the influence of gravity alone. The acceleration due to gravity ('g') is constant for all objects, regardless of their mass, size, or shape.

  7. Which one of the following statements for an object falling freely under the influence of gravity is correct?

    Asked in: UPSC CAPF 2020

    1. Zero acceleration always implies zero velocity
    2. Zero acceleration has no relation with the velocity of the object
    3. Zero velocity at any instant necessarily means zero acceleration at that instant
    4. Acceleration is constant all throughout the free fall
    Show answer

    Answer: D. Acceleration is constant all throughout the free fall

    An object falling freely under the influence of gravity experiences a constant acceleration equal to the acceleration due to gravity (g), assuming air resistance is negligible. This acceleration acts downwards.

  8. The velocity below which a rocket will not escape earth's gravitational pull is - 12 km/s

    Asked in: ASSISTANT ENGINEER(ELECTRICAL) 2016

    1. 9.5 km/s
    2. 11.2 km/s
    3. 13 km/s
    4. 24.25 km/s
    Show answer

    Answer: B. 11.2 km/s

    Escape velocity is the minimum speed an object needs to overcome the gravitational pull of a celestial body and escape into outer space without further propulsion. For Earth, this value is approximately 11.2 kilometers per second.

  9. Why does a ball thrown upwards come down after reaching a certain height?

    Asked in: JUNIOR laboratory assistant 2025

    1. The ball loses its energy
    2. Air pulls it down
    3. Gravity acts on it continuously
    4. It becomes heavier at height
    Show answer

    Answer: C. Gravity acts on it continuously

    This question is a fundamental concept in physics, explaining the action of gravity on objects.

Maximum height and time of flight for vertically projected objects

  1. A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately

    Asked in: UPSC NDA 1 2022

    1. 2 s
    2. 3 s
    3. 4 s
    4. 5 s
    Show answer

    Answer: C. 4 s

    When an object is thrown vertically upward, its velocity decreases due to the acceleration due to gravity ($g$) acting downwards. At the maximum height, the velocity of the object momentarily becomes zero. We can use kinematic equations to find the time taken.

  2. A ball is thrown vertically upward from the ground with a speed of 25-2 m/s. The ball will reach the highest point of its journey in

    Asked in: UPSC NDA 2 2016

    1. 5·14 s
    2. 3·57 s
    3. 2·57 s
    4. 1·29 s
    Show answer

    Answer: C. 2·57 s

    Given initial velocity $u = 25.2 \text{ m/s}$ upwards, and acceleration due to gravity $a = -9.8 \text{ m/s}^2$ (assuming upward is positive). At the highest point, the velocity $v = 0 \text{ m/s}$.

  3. A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of

    Asked in: UPSC NDA 2 2021

    1. 8 m/s
    2. 12 m/s
    3. 16 m/s
    4. 20 m/s
    Show answer

    Answer: D. 20 m/s

    At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g \approx 10 \, \text{m/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) \implies 0 = u^2 - 400 \implies u^2 = 400 \implies u = 20 \, \text{m/s} $. Using $g \approx 9.8 \, \text{m/s}^2$: $0^2 = u^2 + 2(-9.8)(20) \implies 0 = u^2 - 392 \implies u^2 = 392 \implies u \approx 19.8 \, \text{m/s}$. Both approximations are closest to 20 m/s among the given options.

  4. ™Ǒदि ͩकसिी प्रक्षेप्™ का उड्ड™न काल 10 सिेकण्ड हो, तो इसिके द्वारा प्राप्त महत्तम ऊँ चाई क्™ा होगी ? Ǒदि™ा है : g = 10 मीटर/सिेकण्ड²

    Asked in: Assistant Prosecuting Officer (APO) Exam 2021

    1. 100 मीटर
    2. 125 मीटर
    3. 150 मीटर
    4. 200 मीटर
    Show answer

    Answer: B. 125 मीटर

    The maximum height of a projectile can be calculated using its time of flight and acceleration due to gravity. The formula is H = T²g / 8.

Projectile motion: horizontal projection from a height

  1. A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s² and neglecting air resistance will give:

    Asked in: UPSC NDA 1 2023

    1. R=12m
    2. R=18m
    3. R=24m
    4. R=30m
    Show answer

    Answer: C. R=24m

    For projectile motion, the horizontal and vertical components of motion are independent (neglecting air resistance). The time it takes for the stone to hit the ground is determined by the vertical distance and the acceleration due to gravity. The horizontal distance (range) is then calculated by multiplying the horizontal velocity by the time of flight.

  2. A bomb is dropped at a point from a moving aeroplane. The pilot observes that

    Asked in: WBPSC CS Prelims 2019

    1. the bomb traverses a curved path and falls some distance behind that point.
    2. the bomb traverses a curved path and falls some distance ahead.
    3. the bomb drops vertically downward.
    4. the bomb remains stationary in the air for some time.
    Show answer

    Answer: C. the bomb drops vertically downward.

    Neglecting air resistance, an object dropped from a moving vehicle retains the horizontal velocity of the vehicle at the moment of release. Relative to the moving vehicle (the pilot's frame of reference), the object only has a vertical velocity component due to gravity.

  3. Two bullets are fired simultaneously and horizontally from the same vertical height. The first bullet has mass and velocity m₁ and v₁, respectively and the second one has mass and velocity m₂ and v₂ respectively. Which bullet will hit the ground first?

    Asked in: WBCS Main Paper 4 2021

    1. The faster one
    2. The slower one
    3. Both will reach simultaneously
    4. Will depend on masses
    Show answer

    Answer: C. Both will reach simultaneously

    The time it takes for an object launched horizontally to hit the ground depends only on the initial vertical velocity (which is zero) and the vertical distance (height) it falls, and the acceleration due to gravity. Since both bullets are fired from the same height and have the same initial vertical velocity (zero), they will take the same time to reach the ground.

Projectile motion: projection at an angle to the horizontal

  1. A hunter aims his gun at a point between the eyebrows of a monkey sitting on a branch of a tree. Just as he fires, the monkey jumps down. The bullet will

    Asked in: UPPCS Prelims 1999

    1. hit the monkey at the point aimed
    2. hit the monkey below the point aimed
    3. hit the monkey above the point aimed
    4. miss the monkey altogether
    Show answer

    Answer: B. hit the monkey below the point aimed

    This is a classic physics problem demonstrating the independent effects of horizontal motion and vertical motion under gravity. Both the bullet and the monkey are subject to the same downward acceleration due to gravity from the moment the bullet is fired and the monkey drops.

Range, maximum height, and time of flight for angled projectile motion

  1. [Diagram shown in original text] A boy standing at the point ‘O’ in the given diagram, throws a ball three times with the same force, but projecting it along different inclinations from the ground. The results of the throws have been plotted in the diagram. Which one of the following is a valid conclusion?

    Asked in: UPPCS Prelims 1997

    1. The larger the initial inclination, the longer the throw
    2. The larger the height reached, the longer the throw
    3. The larger the height reached, the shorter the throw
    4. The larger the initial inclination, the greater the height reached
    Show answer

    Answer: D. The larger the initial inclination, the greater the height reached

    For a fixed initial speed, the maximum height reached by a projectile is proportional to the square of the sine of the launch angle ($H = \frac{v_0^2 \sin^2\theta}{2g}$). This means that as the launch angle increases from 0 to 90 degrees, the maximum height increases.

  2. A cricket ball is thrown at speed of 28 ms⁻¹ in direction 30° above the horizontal. Calculate the time taken by the ball to return the same level.

    Asked in: ae 2024

    1. 3.5 s
    2. 2.9 s
    3. 2.0 s
    4. 2.2 s
    Show answer

    Answer: A. 3.5 s

    This question involves projectile motion, a fundamental concept in physics. It requires the application of kinematic equations to determine the time of flight.

  3. A ball is thrown at speed of 80 m/s in a direction 30° above the horizontal. The time taken by the ball to return to the same level is (Take, g = 10m/s²)

    Asked in: Assistant Town Planner (Pre.) Exam 2023

    1. 6 s
    2. 8 s
    3. 4 s
    4. 2 s
    Show answer

    Answer: B. 8 s

    The time of flight for a projectile returning to the same horizontal level is determined by the initial vertical velocity component and the acceleration due to gravity. The formula $T = \frac{2 v_0 \sin \theta}{g}$ is essential for solving such problems.

Variation of 'g' with altitude and depth

  1. Statement-I: A body weighs less on a hill top than on earth's surface even though its mass remains unchanged. Statement-II: The acceleration due to gravity of the earth decreases with height.

    Asked in: UPSC NDA 1 2015

    1. Both the statements are individually true and Statement II is the correct explanation of Statement I
    2. Both the statements are individually true but Statement II is not the correct explanation of Statement I
    3. Statement I is true but Statement II is false
    4. Statement I is false but Statement II is true
    Show answer

    Answer: A. Both the statements are individually true and Statement II is the correct explanation of Statement I

    Statement I is true because weight (W) is defined as mass (m) multiplied by the acceleration due to gravity (g), W = mg. A body's mass remains constant regardless of location. Statement II is true because the acceleration due to gravity (g) decreases with increasing height above the Earth's surface. The gravitational force, and hence 'g', depends on the distance from the center of the Earth. At a higher altitude like a hill top, this distance is greater than on the surface at sea level. The relationship is approximately g' = g(R/(R+h))², where R is Earth's radius and h is height. As h increases, g' decreases.

  2. The acceleration due to gravity on the surface of the Earth is maximum and it

    Asked in: UPSC CAPF 2011

    1. increases as we go up
    2. decreases as we go up or down
    3. increases as we go down
    4. neither increases nor decreases as we go up or down
    Show answer

    Answer: B. decreases as we go up or down

    The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM/r². On the surface, r is the Earth's radius (R), so g_surface = GM/R². As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r/r², where M_r is the mass within radius r). Gravity is zero at the Earth's center.

  3. Which one of the following statements is correct ?

    Asked in: UPSC CAPF 2016

    1. Acceleration due to gravity decreases with the increase of altitude
    2. Acceleration due to gravity increases with the increase of depth (assuming earth to be a sphere of uniform density)
    3. Acceleration due to gravity decreases with the increase of latitude
    4. Acceleration due to gravity is independent of the mass of the earth
    Show answer

    Answer: A. Acceleration due to gravity decreases with the increase of altitude

    Understanding how the acceleration due to gravity varies with altitude, depth, latitude, and the mass of the celestial body is essential.

Vertical motion: upward and downward projection of objects

  1. Two balls, A and B, are thrown simultaneously. A vertically upward with a speed of 20 m/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9.8 m/s²?

    Asked in: UPSC NDA 2 2016

    1. The balls will collide after 3s at a height of 30·2 m from the ground
    2. The balls will collide after 2s at a height of 20·1 m from the ground
    3. The balls will collide after 1s at a height of 15·1 m from the ground
    4. The balls will collide after 5s at a height of 20 m from the ground
    Show answer

    Answer: C. The balls will collide after 1s at a height of 15·1 m from the ground

    - Use equations of motion under constant acceleration (gravity). - Define a consistent coordinate system (origin and positive direction). - Set the positions of the two objects equal to find the time of collision. - Use the time of collision to find the position (height) of collision.