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Question

A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff in the upward direction with a velocity of 100 ms−1. At what time will the bullet and the block meet ?

A
4s
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B
3s
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C
2s
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D
1s
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Solution

The correct option is D 1s
Lets1 and s2 be the displacements covered by wooden block and bullet respectively.
Given. height of the tower is 100 m
Therefore, s=s1+s2=100 m.
From the second equation of motion,
s=ut+12×at2
Before the collision, in the case of block, initial velocity,u = 0 and the acceleration, a = g (acceleration due to gravity)
s1=12gt2=12×9.8×t2=4.9t2
In the case of bullet, initial velocity, u=100 ms1 and the acceleration, a = g (acceleration due to gravity)
s2=ut12gt2
=100t12×9.8×t2=100t4.9t2
Substituting the values in equation, s1+s2=100 m, we get,
4.9t2+100t4.9t2=100
100t = 100
t = 1 s.

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