A bullet on penetrating 30 cm into its target loses it's velocity by 50%. What additional distance will it penetrate into the target before it comes to rest?

- Let the initial velocity of the bullet be v₀ and the constant deceleration be 'a' (negative value). - In the first 30 cm (x₁ = 0.3 m), the velocity reduces by 50%, meaning the velocity becomes v₁ = v₀/2. - Using the kinematic equation v² = u² + 2as: (v₀/2)² = v₀² + 2a(0.3). - This simplifies to v₀²/4 = v₀² + 0.6a, which gives -3v₀²/4 = 0.6a, or a = -3v₀² / (4 0.6) = -3v₀² / 2.4. - For the second part, the bullet starts with velocity v₁ = v₀/2 and comes to rest (v₂ = 0) after penetrating an additional distance x₂. - Using the same kinematic equation: 0² = (v₀/2)² + 2a(x₂). - This simplifies to 0 = v₀²/4 + 2ax₂. Substituting the value of 'a': 0 = v₀²/4 + 2 (-3v₀²/2.4) x₂. - 0 = v₀²/4 - (assuming v₀ ≠ 0): 1/4 = (6/2.4) x₂ => 1/4 = (60/24) x₂ => 1/4 = (5/2) x₂. - Solving for x₂: x₂ = (1/4) (2/5) = 2/20 = 1/10 m = 10 cm.