Aball is thrown horizontally from the top of a tall cliff. neglecting air drag, what vertical distance will the ball have fallen after 3 seconds?

The relevant equation to use here is:

y = v0 t + 0.5 g t^2

where y is the vertical distance, v0 is initial velocity = 0, t is time, g = 9.8 m/s^2

y = 0 + 0.5 * 9.8 * 3^2

y = 44.1 meters

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