A stone is dropped from rest from the top of a building. During its fall it crosses a vertical window of height 2 m in 0.20 s. Neglect air resistance and take g = 10 m/ $s^2.$ The distance of the top edge of the window below the point from which the stone was dropped is:
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Question

A stone is dropped from rest from the top of a building. During its fall it crosses a vertical window of height 2 m in 0.20 s. Neglect air resistance and take g = 10 m/ $s^2.$ The distance of the top edge of the window below the point from which the stone was dropped is:

Options

G1 Physics · Accepted Answer

Let v be the speed at the top edge of the window. For crossing the 2 m window in 0.20 s: 2 = v(0.20) + (1/2)(10)(0.20)^2 = 0.2v + 0.2, so v = 9 m/s. From the release point to the top of the window, v^2 = 2gh. Thus h = 81/(20) = 4.05 m. Hence option B is correct.

A stone is dropped from rest from the top of a building. During its fall it crosses a vertical window of height 2 m in 0.20 s. Neglect air resistance and take g = 10 m/s^2. The distance of the top edge of the window below the point from which the stone was dropped is: