1. Motion of the bullet before hitting the ground

The bullet is projected vertically downward with initial velocity $u = 100\ \text{m/s}$ and acceleration due to gravity $g = 10\ \text{m/s}^2$ (downward).

Taking downward direction as positive, its velocity after time $t$ is $v = u + gt = 100 + 10t.$

So, from $t=0$ until it hits the ground, the velocity increases linearly with time. That means the $v$-$t$ graph is a straight line with positive slope $10$ and intercept $100$.

2. Condition of hitting the ground within 10 s

The question says the bullet reaches the ground within 10 s. Let the hitting time be $t_1$, where $t_1 < 10\ \text{s}.$

So the straight-line increase in velocity continues only up to some time less than $10$ s.

At that instant, the bullet collides with the ground.

3. Effect of perfectly inelastic collision

The bullet "instantaneously comes to rest" on hitting the ground. That means at $t=t_1$, velocity suddenly changes from $v = 100 + 10t_1$ to $v = 0.$

So the graph must show a vertical drop from a positive velocity to zero at the instant of collision.

4. Motion after collision

After collision, the bullet remains at rest on the ground. So for the remaining time up to $20$ s, $v=0.$

Hence from $t=t_1$ to $t=20$ s, the graph is a horizontal line along the time axis.

5. Required shape of the graph

Therefore, the correct $v$-$t$ graph must have:

• initial velocity $100\ \text{m/s}$ at $t=0$,
• a straight line increasing with slope $+10\ \text{m/s}^2$,
• collision occurring at some time $<10$ s,
• an instantaneous vertical fall to zero,
• then velocity remaining zero till $t=20$ s.

This corresponds to Option A.

6. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.