A driver running 1.9 m/s dives out horizontally from the edge of a vertical clif

Question

A driver running 1.9 m/s dives out horizontally from the edge of a vertical cliff and 2.4 s later reaches the water below. Part A) How high was the cliff?
Part B) How far from its base did the diver hit the water? A driver running 1.9 m/s dives out horizontally from the edge of a vertical cliff and 2.4 s later reaches the water below. Part A) How high was the cliff?
Part B) How far from its base did the diver hit the water? Part A) How high was the cliff?
Part B) How far from its base did the diver hit the water?

Explanation / Answer

vertically

S = ut + 1/2 at2

S = 0+ 0.5*9.81*2.4^2 = 28.2528m high

Horizontally

s = ut +1/2 at2

Far = 1.9*2.4 = 4.56m

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