A particle is projected vertically upward from the ground with speed u. At the same instant another particle is dropped from rest from a height equal to the maximum height of the first particle. They meet at:
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Question

A particle is projected vertically upward from the ground with speed u. At the same instant another particle is dropped from rest from a height equal to the maximum height of the first particle. They meet at:

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G1 Physics · Accepted Answer

Maximum height of the projected particle is H = u²/(2g). Both particles have the same acceleration −g, so their relative acceleration is zero. Initial relative separation = H. Relative speed = u. Thus t = H/u = u/(2g). Height of the projected particle at this time: h = ut − gt²/2 = u[u/(2g)] − (g/2)[u²/(4g²)] = u²/(2g) − u²/(8g) = 3u²/(8g).