A balloon moves vertically upward with constant speed V. Stone A is released from it. After a time τ, stone B is released from the same balloon. At time t after release of B, the separation between the two stones is:
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Question

A balloon moves vertically upward with constant speed V. Stone A is released from it. After a time τ, stone B is released from the same balloon. At time t after release of B, the separation between the two stones is:

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G1 Physics · Accepted Answer

Take the release point of A as y=0. At time τ+t, stone A is at position of A = V(τ+t) − g(τ+t)²/2. Stone B was released from the balloon at height Vτ and has initial velocity V: position of B = Vτ + Vt − gt²/2 = V(τ+t) − gt²/2. Therefore position of B−position of A = (g/2)[(τ+t)²−t²] = gτ(t+τ/2).