Question

A ball is dropped vertically from a height $10\text{m}$ above the ground. It hits the ground and bounces up vertically to a height of $5\text{m}$. Neglecting subsequent motion and air resistance, which of the following options correctly represents the variation of its velocity $v$ with height $h$ above the ground.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let $h_0$ be the height of the ball from the ground at time $t=0$ seconds.
Given that, $\left(h_0\right)=10\text{m}$
Initial velocity $\left(u\right)=0\text{m/s}$


We know that for a uniformly accelerated motion under gravity
$v^2-u^2=2g(h_0-h)$
$\Rightarrow v^2=2g(10-h)----\left(1\right)$
[$\because u=0$]
Let us assume that, the direction of upward motion is positive and direction of downward motion is negative as shown in the figure.
And from (1), we can say that as height decreases $(h\downarrow )$, velocity of ball increases $(v\uparrow )$ in negative direction. Velocity reaches its maximum when the height becomes zero.
After hitting the ground, the ball reverses its direction and as height increases $(h\uparrow )$ there is decrease in velocity $(v\downarrow )$ in positive direction.
The relation between velocity and height is parabolic in nature, which is similar to the equation of a parabola $y^2=-ax+b$
This variation is shown only by option (a).
Hence, option (a) is the correct answer.