Question:
A stone is projected vertically upward from some height at 2v. Some ∆t later another stone is projected vertically upward at v from the same position. If they reach the initial position at the same time, find v.
Answer:
Time taken by the stone to reach at initial position = Time of flight (T) of stone = Time of ascent(tA) +Time of descent(tD)
tA= tD = u/g (u is initial velocity)
So, T = 2u/g
For first stone :
Initial velocity = 2v
So, T = 2(2v)/g = 4v/g
For second stone:
u= v
So, T = 2v/g
Given that the second stone is projected after Δ t time.
So, 4v/g = 2v/g +Δ t
Hence, v = Δtg/2
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