Question

A projectile moving vertically upwards with a velocity of $200m{s}^{-1}$ breaks into two equal parts at a height of $490m$. One part starts moving vertically upwards with a velocity of $400m{s}^{-1}$. How much time it will take, after the break up with the other part to hit the ground?

• A. $2\sqrt{10}S$
• B. $5s$
• C. $10s$
• D. $\sqrt{10}S$

Answer 1

Answer: (C)

Momentum before explosion
=Momentum after explosion
$m\times 200\hat{j}=\frac{m}{2}\times 400\hat{j}+\frac{m}{2}v$
$=\frac{m}{2}\left(400\hat{j}+v\right)$
$\Rightarrow 400\hat{j}-400\hat{j}=v$
$\therefore v=0$
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, $h=ut+\frac{1}{2}g{t}^2$
$490=0+\frac{1}{2}\times 9.8\times t^2$
$\Rightarrow t^2=\frac{980}{9.8}=100$
$\therefore t=\sqrt{100}=10sec$