Q 11-02-099JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
A bullet is shot vertically downwards with an initial velocity of $100\ \text{m s}^{-1}$ from a certain height. Within $10$ s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time $t = 20$ s will be : (Take $g = 10\ \text{m s}^{-2}$)
Answer: (A) see figure
Take upward as positive. Then $u = -100\ \text{m s}^{-1}$ and $a = -10\ \text{m s}^{-2}$:
$$v = -100 - 10t$$
The velocity goes linearly from $-100$ to $-200\ \text{m s}^{-1}$ in the first $10$ s. The bullet then stops at once, so $v$ jumps to $0$ and stays $0$ until $t = 20$ s. This is graph (1).
Solution by Sreeraj P, M.Sc Physics