A projectile moving vertically upwards with a velocity of 200 ~ms^-1 breaks into two equal parts at a height of 490 ~m. One part starts moving…
JEE Main 2012 — Physics Mechanics
2012mcqmedium
A projectile moving vertically upwards with a velocity of 200ms−1 breaks into two equal parts at a height of 490m. One part starts moving vertically upwards with a velocity of 400ms−1. How much time it will take, after the break up with the other part to hit the ground?
Official previous-year question
Held on 12 May 2012 · Verified 6 Jul 2026.
Options
A
210s
B
5s
C
10s
D
10s
Did you get this right?
Sign in to track your attempts and accuracy.
Solution
Momentum before explosion = Momentum after explosion m×200j^=2m×400j^+2mv=2m(400j^+v)⇒400j^−400j^=v∴v=0 i.e., the velocity of the other part of the mass, v=0 Let time taken to reach the earth by this part be t Applying formula, h=ut+21gt2490=0+21×9.8×t2⇒t2=9.8980=100∴t=100=10sec
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.