Particle A is projected vertically upward from the ground at t=0 with speed u. At t=τ, particle B is released from rest from height H. Find the minimum value of u such that A meets B before or exactly at the highest point of A. Assume τ<√(2H/g).
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Question

Particle A is projected vertically upward from the ground at t=0 with speed u. At t=τ, particle B is released from rest from height H. Find the minimum value of u such that A meets B before or exactly at the highest point of A. Assume τ<√(2H/g).

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G1 Physics · Accepted Answer

For t≥τ, position of A = ut − gt²/2, position of B = H − g(t−τ)²/2. Because both particles have the same acceleration after B is released, their separation is linear in t: position of B−position of A = H − gτ²/2 − (u−gτ)t. Thus collision time is collision time = [H−gτ²/2]/(u−gτ). For the minimum projection speed that still allows collision while A is ascending, the limiting collision occurs exactly at A's highest point: collision time = u/g. Therefore [H−gτ²/2]/(u−gτ) = u/g. This gives u² − gτu…