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

Options

A.
B.
C.
D.

Answer

A

Explanation

(i) for uniformly acceleration/ decelerated 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 (A) satisfies both these condition. Therefore, correct answer is (A). Note the time `t=0` corresponds to the point on the graph where `h=d`