A ball is thrown vertically upward from the ground at 30 m s⁻¹. One second later, another ball is thrown vertically upward from the same point at 40 m s⁻¹. Taking g = 10 m s⁻², the two balls meet after the first ball was thrown at:
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Question

A ball is thrown vertically upward from the ground at 30 m s⁻¹. One second later, another ball is thrown vertically upward from the same point at 40 m s⁻¹. Taking g = 10 m s⁻², the two balls meet after the first ball was thrown at:

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G1 Physics · Accepted Answer

For the first ball: y₁ = 30t − 5t². For the second, for t ≥ 1: y₂ = 40(t − 1) − 5(t − 1)². Equating and simplifying gives 30t − 5t² = 40t − 40 − 5t² + 10t − 5 = 50t − 45 − 5t². Thus 20t = 45, so t = 2.25 s.