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

When a ball is falling down from height (d) and it reaches at height (h) from ground, then the velocity acquired by the ball is `
`v = sqrt 2 ( g (d- h)) ` or ` v^2 =2 g (d- h)` ….(i)`
Thuis downward velocity of ball is begative.
` When ` h=d, v=0` and when ` h=0, v= sqrt 2 g d`
From (i), ` v` adn `h` variation is a parabola below h-axis.
After collsision with ground , the ball can go vertically upward upto a height ` d//2` , so its initial upward velocity is
` ` u' = sqrt (2 g (d//2)` . At height (h) its velocity is
` v' = sqrt( u'^2-2 gh) = sqrt( 2 g (d//2)-2 gt)` ...(ii)`
When , ` h= d//2, v' =0`. The upward velocity of ball is positive.
From (ii) the variation of `v'` and `h` is a parabola above h-axis. Thus graph (a) satisfies all these conditions.