Asked by Lakisha
A runner runs and jumps off a cliff into the water below. If the runner takes up running a 9.37 m/s and lands in the water 2.0 m away from the cliff's edge: How tall is the cliff?
Answers
Answered by
drwls
The time it takes to hit the water is
t = 2.0/9.37 = 0.213 s
Now calculate how far he falls in that time.
Y = (g/2) t^2
That will be the cliff height.
t = 2.0/9.37 = 0.213 s
Now calculate how far he falls in that time.
Y = (g/2) t^2
That will be the cliff height.
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