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Question

A body A is thrown vertically upward with the initial velocity v1. Another body B is dropped from a height h. Find how the distance x between the bodies depends on the time t if the body begin to move simultaneously.


A
x=(hv1)t
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B
x=hv1t
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C
x=hv1t
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D
x=htv1
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Solution

The correct option is B x=hv1t
Let h1 and h2 be the distance travelled by the body A and B respectively in time t, and x be the distance between the bodies.
Then, we have
h1=v1tgt22 and h2=gt22....(i)
So, we have distance between the bodies
x=h(h1+h2)
From (i), h1+h2=v1t
Hence, we have
x=hv1t

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