Animated Solution for Physics - Kinematics: A ball is dropped vertically from a height $d$ above the ground. It hits the ground and bounces up vertically to a height $\frac{d}{2}$. Neglecting subsequent motion and air resistance, its velocity $v$ varies with height $h$ above the ground as

Decoding the $v-h$ Graph

Imagine standing on a balcony and dropping a ball. It accelerates downwards, smashes into the ground, and rebounds, but it doesn't quite reach your hand again—it only makes it halfway up.

This is a classic kinematics problem, but with a twist: we aren't plotting velocity against time, but velocity against height ($h$). This subtle change in variables completely transforms the geometry of our graphs.

Since time ($t$) is completely absent from our axes, we need a mathematical tool that bypasses it entirely. Enter the third equation of motion:

This equation is our golden key because it directly links velocity, acceleration, and displacement. Let's break the ball's journey into two distinct phases and apply this equation to each.

The Downward Fall (The Negative Parabola)

During the fall, the ball starts from rest at a height $h=d$. Therefore, its initial velocity is $u=0$. The acceleration acting on it is gravity, pulling it downwards, so $a=-g$. The displacement from the starting point to any intermediate height $h$ is simply final position minus initial position, or $s=h-d$.

Plugging these into our master equation, we get:

Rearranging this gives us the equation for the first phase:

Notice the mathematical structure here. The velocity $v$ is squared, while the height $h$ is linear. In coordinate geometry, an equation of the form $y^2\propto -x$ represents a parabola opening towards the negative x-axis.

However, we must be incredibly careful with our sign convention. Because the ball is physically moving downwards, its velocity vector points in the negative direction. Therefore, when we take the square root, we must select the negative root:

This gives us the lower half of a parabola in the fourth quadrant of our $v-h$ graph, starting at $v=0$ when $h=d$, and reaching a maximum downward speed of $-\sqrt{2gd}$ just before it hits the ground at $h=0$.

When the ball strikes the ground at $h=0$, a dramatic event occurs. In our idealized physics model, the collision happens instantaneously. The ball's velocity abruptly changes from a large negative value (moving down) to a positive value (moving up) without any change in height.

On a graph, an instantaneous change in the y-variable (velocity) without a change in the x-variable (height) manifests as a perfectly vertical line. This represents the sudden impulse force from the ground reversing the ball's momentum.

The Upward Bounce (The Positive Parabola)

Now the ball is moving upwards. We are told it reaches a maximum height of $\frac{d}{2}$. At this peak, its final velocity $v_f$ becomes zero. The acceleration remains $a=-g$.

Let's apply our master equation again, this time from an arbitrary height $h$ during the bounce up to the peak at $\frac{d}{2}$:

Solving for $v^2$, we find:

Once again, we have a parabolic relationship. But this time, the ball is moving upwards, so its velocity is positive. We take the positive square root:

This represents the upper half of a parabola in the first quadrant. It starts at a positive velocity of $+\sqrt{gd}$ right after the bounce at $h=0$, and smoothly curves down to $v=0$ at $h=\frac{d}{2}$.

By piecing together the mathematical realities of both phases, we are looking for a graph that shows a negative parabolic arc from $h=d$ to $h=0$, a vertical jump at $h=0$, and a smaller positive parabolic arc from $h=0$ to $h=\frac{d}{2}$.

Looking at the given options, Graph (a) is the only one that flawlessly captures this beautiful parabolic dance!