AP Physics Question

Question

It takes .21s for a dropped object to pass a window that is 1.35m tall. From what height above the top of the window was the object released?

Answer 1

One straightforward way is to use d = 1/2gt² for how far something falls in a given time.

height of window = distance fallen fin time t + .21 - distance fallen in time t where t is the time to reach top of the window.

h = 1/2 g(t+.21)² - 1/2 gt² = g(.21)t - (1/2)g(.21)²

1.35 = (9.8)(.21)t - 4.9(.21)²

Solve for t and then plug into d = 1/2 gt² to obtain d dropped to reach top of the window.

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