Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. she starts with a velocity of 4.00m/s4.00m/s and her takeoff point is 8m8m above the pool. how long are her feet in the air?

Answer 1

First we can find the swimmer's velocity when the feet enter the water. v^2 = (v0)^2 + 2gy v = sqrt{ (v0)^2 + 2gy } v = sqrt{ (4.00 m/s)^2 + (2)(-9.80 ~m/s^2)(-8 m) } v = sqrt{ 172.8 } v = -13.145 m/s Note that the negative sign means the swimmer is moving downward. We can find the time to reach this velocity. t = (v - v0) / g t = (-13.145 m/s - 4.00 m/s)) / -9.80 m/s^2 t = - 17.145 m/s / -9.80 m/s^2 t = 1.75 seconds Her feet are in the air for 1.75 seconds.