A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d 2 , Neglecting subsequent motion and air resistance, its velocity v varies with height h above the ground as

Step-by-step solution

(i) For uniformly accelerated/deaccelerated motion, v 2 = u 2 ± 2 g h i.e., v - h graph will be a parabola (because the equation is quadratic). (ii) Initially, velocity is downwards (-ve) and then after the collision it reverses its direction with lesser magnitude. i.e., velocity is upwards (+ve). Graph (a) satisfies both these conditions. Note that time t = 0 corresponds to the point on the graph where h = d .