Physics Circle home Student Login

Home›Chapters›Motion in One Dimension PYQs›Motion Under Gravity›Question 29

JEE Main 2022 (27 Jul, Shift 1) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2022 (27 Jul, Shift 1)

Q29JEE Main 2022 · 27 JUL · SHIFT 1Medium

Motion Under Gravity

A bullet is shot vertically downwards with an initial velocity of 100 m/s100\text{ m/s} 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 st = 20\text{ s} will be:

(Take g = 10 m/s2).

Show answer and solution

Answer: (a)

  1. Take upward as positive: v=−100−10tv = -100 - 10t
  2. At t=10 st = 10\text{ s}: v=−200 m/sv = -200\text{ m/s}, a straight line from −100-100 to −200-200.
  3. Perfectly inelastic hit: velocity drops to zero at once and stays zero up to 20 s.
  4. So the answer is (a).
Report a mistake

More questions on Motion Under Gravity