A freely falling body crosses an upper window of height 30 m in 2 s and a lower window of the same height in 1 s. Taking g=10 m s⁻², find the vertical separation between the top edges of the two windows, in metre.
Physics · Kinematics Part-1 (one dimension) · Motion under gravity
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Question

A freely falling body crosses an upper window of height 30 m in 2 s and a lower window of the same height in 1 s. Taking g=10 m s⁻², find the vertical separation between the top edges of the two windows, in metre.

Physics · Kinematics Part-1 (one dimension) · Motion under gravity

G1 Physics · Accepted Answer

For the upper window: 30 = v₁(2) + 5(2²) ⇒ 30 = 2v₁+20 ⇒ v₁=5 m s⁻¹. For the lower window: 30 = v₂(1) + 5 ⇒ v₂=25 m s⁻¹. If S is the separation between the top edges, v₂² = v₁² + 2gS. Thus S = (25²−5²)/(20) = 600/20 = 30 m.