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 the height h above the ground as
Text Solution
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(i) For uniformly accelerated/deaccelerated motion
v 2 = u 2 ± 2gh
i.e. v-h graph will be a parabola (because equation is quadratic).
(ii) Initially velocity is downwards (-ve) and then after collision it reverses its direction with lesser magnitude. i.e. velocity is upwards (+ve). Graph satisfies both these conditions.
Therefore, correct answer is
Note that time t = 0 corresponds to the point on the graph where h = d
Next time collision takes place at 3.
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