As tennis ball is dropped, so its initial velocity, u = 0

Suppose its velocity at any time t = v

Acceleration due to gravity = g

$$\therefore$$ v = u + gt

$$\Rightarrow$$ v = gt ..... (i)

We know, kinetic energy,

$$K = {1 \over 2}m{v^2} = {1 \over 2}m({g^2}{t^2})$$ [Using eqn. (i)]

$$ \Rightarrow K = {1 \over 2}m{g^2}{t^2}$$

$$ \Rightarrow K \propto {t^2}$$ (As m and g are constant)

This shows that relation between kinetic energy of tennis ball K and time t is parabolic.

During the collision, velocity of the ball falls sharply to zero as it is compressed and regains maximum velocity in the same short time interval.

This relation is best illustrated by the option (b).