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 bodies begin to move simultaneously.
A
x=h−v1t
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B
x=(h−v1)t
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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 Ax=h−v1t Let the distance covered by body A be h1 ⇒h1=v1t−12gt2 and, let the distance travelled by the body B be h2 ⇒h2=0+12gt2 Given, the distance between the bodies is x=h−(h1+h2) ∵h1+h2=v1t ⇒x=h−v1t Hence, the distance x between the bodies depends on the time t if the bodies begin to move simultaneously as x=h−v1t