• # question_answer A projectile moving vertically upwards with a velocity of 200 $\text{m}{{\text{s}}^{-1}}$breaks into two equal parts at a height of 490 m. One part starts moving vertically upwards with a velocity of 400 $\text{m}{{\text{s}}^{-1}}$.                    [JEE ONLINE 12-05-2012] How much time it will take, after the break up with the other part to hit the ground? A) $2\sqrt{10}s$ B) 5 s C) 10 s D) $\sqrt{10}s$

 [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$ $\therefore$ $t=\sqrt{100}=10\sec$