A rock is kicked horizontally off the edge of the cliff at a velocity of +15.0 m/s. If the canyon below is 80.0 m deep, how far from the edge of the cliff does the model rocket land?

60.6 m

Explanation:

For vertical displacement of 80 m , initial velocity downwards u = 0

h = ut + 1/2 g t²

80 = 0 + 1/2 x 9.8 x t²

t = 4.04 s

During this period , rock will be displaced horizontally with uniform velocity .

horizontal displacement = 15 x 4.04 = 60.6 m .

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