WAEC Physics 2015 Paper

Concept:For vertical upward motion, the velocity at maximum height becomes zero, and gravity provides the deceleration.Use the equation of motion $v^2 = u^2 + 2as$.Explanation:Let $u$ be the initial velocity of the bullet.At maximum height $h = 500\,m$, the final velocity is $v = 0$.Since the bullet moves upward against gravity, the acceleration is $a = -g$, where $g = 10\,m\,s^{-2}$.Substitute into $v^2 = u^2 + 2as$:$0 = u^2 - 2gh$$u^2 = 2gh$$u^2 = 2 \times 10 \times 500$$u^2 = 10{,}000$$u = \sqrt{10000} = 100\,m\,s^{-1}$Therefore, the magnitude of the initial velocity is $100\,m\,s^{-1}$.Answer:$100.0\,m\,s^{-1}$ Option B.