A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below

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A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below

A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below. How high was the cliff, and how far from its base did the diver hit the water?

physics physic

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Answer 1

Answer #1

h = ½gt², since diver's initial vertical speed is 0
h = 44 m

x = vt
x = 5.4 m

answered by: Demtrius

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Answer 2

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A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below

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A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below

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A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below