Q. 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 vertically upwards with a velocity of $400 \,ms^{-1}$. How much time it will take, after the break up with the other part to hit the ground?
AIEEEAIEEE 2012Work, Energy and Power
Solution:
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\quad\quad\right)$
$\Rightarrow \,400\hat{j}- 400\hat{j} = v$
$\therefore \quad v=0$
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, $h = ut + \frac{1}{2}gt^{2}$
$490 = 0+\frac{1}{2}\times9.8\times t^{2}$
$\Rightarrow \quad t^{2} = \frac{980}{9.8} = 100$
$\therefore \quad t = \sqrt{100} = 10\,sec$
