91Ó°ÊÓ

Velocity denotes the variation in the displacement of the ball with respect to a particular frame of reference and the time taken by the ball.

Acceleration denotes the variation in the velocity of the ball with respect to the required time.

When the ball is thrown upward, it becomes stationary for a very short duration and acquires zero velocity. Then, the ball returns under the effect of gravity.

This way, the distance between the initial and final positions, i.e., displacement, becomes zero.

So, the velocity can be expressed as:

$V=\frac{\mathrm{displacement}}{\mathrm{time}}$

From the above expression, the velocity is zero.

Thus, the velocity at the highest point in the ball’s path is zero.

When the ball moves upward, the acceleration due to gravity acts on it downward. Due to this, the ball returns to the ground.

The acceleration (a) is equal to the acceleration due to gravity (g).

$$\begin{gathered} a=g \\ a=9.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration of the ball at the highest point is $9.8\mathrm{m}/{\mathrm{s}}^2$, acting in the downward direction.

Option (a) is incorrect because the velocity at the top is zero, but the acceleration is non-zero at$9.8\mathrm{m}/{\mathrm{s}}^2$.

Option (b) is incorrect because the velocity at the highest point is zero, but the acceleration is $9.8\mathrm{m}/{\mathrm{s}}^2$, in the downward direction.

Option (d) is incorrect because the velocity at the highest point is zero, while the acceleration is not.

Option (e) is incorrect because the velocity at the topmost point is not $9.8\mathrm{m}/\mathrm{s}$, and the acceleration is also not $0\mathrm{m}/{\mathrm{s}}^2$.

The velocity at the topmost point is zero, and the acceleration is $9.8\mathrm{m}/{\mathrm{s}}^2$, in the downward direction.

Hence, option (c) is correct.