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=(h−v1)t
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B
x=h−v1t
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C
x=h−v1t
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D
x=ht−v1
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Solution
The correct option is Bx=h−v1t 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=v1t−gt22 and h2=gt22....(i)
So, we have distance between the bodies x=h−(h1+h2)
From (i), h1+h2=v1t
Hence, we have x=h−v1t