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a diver running 1.6m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.0s later. how hig...Asked by Al
A diver running 1.2 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.1 s later. How high was the cliff and how far from its base did the diver hit the water?
Answers
Answered by
George
vertical drop
S = ut+(1/2)gt²
= 0 + (1/2)(9.8)*(3.1)^2
=47.089s
How far from base
1.2m/s * 3.1s = 3.72m
S = ut+(1/2)gt²
= 0 + (1/2)(9.8)*(3.1)^2
=47.089s
How far from base
1.2m/s * 3.1s = 3.72m
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