Asked by Yuli
A 62.7-kg person jumps from rest off a 2.98-m-high tower straight down into the water. Neglect air resistance. She comes to rest 1.09 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.
please help :D? i really don't get this question
please help :D? i really don't get this question
Answers
Answered by
Damon
Calculate speed when body hits water:
(1/2) m v^2 = m g h
so
v = sqrt(2 g h)
average force = change in momentum /time to stop
change in momentum = m(v-0)
time to stop = distance/average speed = 1.09 / (.5 v)
(1/2) m v^2 = m g h
so
v = sqrt(2 g h)
average force = change in momentum /time to stop
change in momentum = m(v-0)
time to stop = distance/average speed = 1.09 / (.5 v)
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