91Ó°ÊÓ

The velocity of the dropped ball varies with air resistance on the ball with height. The velocity of the ball increases up to a certain position but becomes constant after this certain position. The constant velocity is called the terminal velocity and it is constant because the air resistance on the ball after the terminal velocity of the ball becomes constant.

The velocity of the falling ball is given by the following equation:

$\mathrm{v}={\mathrm{v}}_{\mathrm{t}}\left(1-{\mathrm{e}}^{-{\left(\mathrm{k}/\mathrm{m}\right)}^{\mathrm{t}}}\right)$ ….. (1)

Here, ${\mathrm{v}}_{\mathrm{t}}$is the terminal speed of the ball, k is drag coefficient, m is the mass of the falling ball, t is the time for falling.

Let the initial acceleration is g.

As speed increases resistive force increases so the net force decreases. Thus acceleration decreases as velocity become maximum which is terminal velocity, and so the resistive force is maximum and balanced the force mg. Therefore net force on the ball becomes zero. And thus acceleration becomes zero after a certain time.

Differentiate equation (1) to find the acceleration

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \\ =\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{v}}_{\mathrm{t}}\left(1-{\mathrm{e}}^{-\left(\mathrm{k}/\mathrm{m}\right)\mathrm{t}}\right)\right] \\ ={\mathrm{ge}}^{-\left(\mathrm{k}/\mathrm{m}\right)\mathrm{t}} \end{gathered}$$

Hence, the acceleration of the falling ball is decreasing exponentially with time so the correct graph for the falling ball will be graph (d).