A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t₁. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t₂. Time required to reach the ground, if it is dropped from the top of the tower, is :
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Question

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t₁. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t₂. Time required to reach the ground, if it is dropped from the top of the tower, is :

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G1 Physics · Accepted Answer

Let tower height be h and initial speed magnitude be u. For upward projection: t₁ = [u + √(u² + 2gh)]/g For downward projection: t₂ = [-u + √(u² + 2gh)]/g Therefore: t₁t₂ = [(u² + 2gh) - u²]/g² = 2h/g If the body is simply dropped: t = √(2h/g) Hence: t = √(t₁t₂)