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, \[h=ut+\frac{1}{2}g{{t}^{2}}\] |
\[490=0+\frac{1}{2}\times 9.8\times {{t}^{2}}\]\[\Rightarrow \]\[{{t}^{2}}=\frac{980}{9.8}=100\] |
\[\therefore \] \[t=\sqrt{100}=10\sec \] |
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