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) \[\sqrt{10}s\]

Correct Answer: C

Solution :

[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, \[h=ut+\frac{1}{2}g{{t}^{2}}\]

\[490=0+\frac{1}{2}\times 9.8\times {{t}^{2}}\]\[\Rightarrow \]\[{{t}^{2}}=\frac{980}{9.8}=100\]

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