A freely falling particle crosses a vertical window of height 20 m in 1 s. Taking g=10 m s⁻², the distance between the point of release and the top edge of the window is:
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A freely falling particle crosses a vertical window of height 20 m in 1 s. Taking g=10 m s⁻², the distance between the point of release and the top edge of the window is:

G1 Physics · Accepted Answer

Let the speed at the top edge of the window be v. During the 1 s crossing: 20 = v(1) + (1/2)(10)(1²) ⇒ v = 15 m s⁻¹. The particle started from rest, so the distance fallen before reaching the top edge is s = v²/(2g) = 225/20 = 11.25 m.