A balloon rises at 5 m s⁻¹. At a height of 100 m, stone A is released. Two seconds later, stone B is thrown vertically downward from the balloon at 15 m s⁻¹ relative to the balloon. Taking g = 10 m s⁻², which reaches the ground first?
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Question

A balloon rises at 5 m s⁻¹. At a height of 100 m, stone A is released. Two seconds later, stone B is thrown vertically downward from the balloon at 15 m s⁻¹ relative to the balloon. Taking g = 10 m s⁻², which reaches the ground first?

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G1 Physics · Accepted Answer

For A: −100 = 5t − 5t², giving t = 5 s. After 2 s the balloon is at 110 m. B has ground velocity 5 − 15 = −10 m s⁻¹. For B: −110 = −10τ − 5τ² ⇒ τ² + 2τ − 22 = 0, so τ ≈ 3.90 s after its throw, or 5.90 s after A was released. Therefore A reaches about 0.90 s earlier, closest to 1.00 s.