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 \[\text{m}{{\text{s}}^{-1}}\]. How much time it will take, after the break up with the other part to hit the ground?   JEE Main Online Paper (Held On 12 May 2012)

A) \[2\sqrt{10}s\]                 

B)                        5 s

C)                        10 s       

D)                        \[\sqrt{10}s\]

Correct Answer: C

Solution :

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