Animated Solution for Physics - Kinematics: Consider a rubber ball freely falling from a height $h=4.9\text{ m}$ onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then, the velocity as a function of time the height as function of time will be

Imagine standing on a balcony and dropping a rubber ball. As soon as it leaves your hand, gravity takes over. The ball accelerates downwards, gaining speed until it smacks into the ground. But this isn't just any floor—it's a perfectly elastic plate. This means the ball doesn't lose any energy when it hits; it bounces back with the exact same speed it had right before the impact.

Our goal is to translate this physical reality into a velocity-time graph. To do this, we first need to establish a strict sign convention. Let's take the upward direction as positive ($+y$) and the downward direction as negative ($-y$).

Since the ball is in free fall, the only force acting on it is gravity. Therefore, its acceleration is constant: $a=-g$.

Using the first equation of motion, $v=u+at$, and knowing the ball starts from rest ($u=0$), we get our master equation for the flight:

This equation tells us two crucial things about the velocity-time graph: 1. The graph must be a straight line. 2. The slope of this line must be constant and negative (specifically, $-g$).

Let's call the time it takes for the ball to hit the ground $t_1$. During this time, the velocity becomes increasingly negative. Just before it hits the ground, its velocity is $-v_1=-g{t}_1$.

On our graph, this phase is represented by a straight line starting from the origin $(0,0)$ and going down to $(t_1,-v_1)$.

Now comes the interesting part. The problem states that the collision duration is negligible and perfectly elastic. In the blink of an eye, the ball's velocity goes from $-v_1$ (moving down) to $+v_1$ (moving up).

Because this happens in zero time, the graph doesn't slope up; it jumps vertically! We draw a dashed vertical line from $-v_1$ straight up to $+v_1$ at $t=t_1$.

After the bounce, the ball is moving upwards with velocity $+v_1$. But gravity is still pulling it down, so the velocity decreases linearly, following the same slope of $-g$.

It takes exactly $t_1$ seconds for gravity to slow the ball down to zero at its highest point. So, the graph crosses the time axis at $t_1+t_1=2t_1$.

From the highest point, it falls back down, taking another $t_1$ seconds to reach the ground. It hits the ground again with velocity $-v_1$ at $t=2t_1+t_1=3t_1$.

Here is the catch: Many students fall for Option (a) because it shows bounces at equal intervals of $t_1$. But they forget that after the first bounce, the ball has to travel up and down, which takes twice as long ($2t_1$) as the initial fall!

Therefore, the bounces occur at $t_1$, $3t_1$, $5t_1$, and so on. Option (c) perfectly captures this extended time interval between bounces, making it the correct choice.