Question
A diver of mass 43.2 kg jumps off a high dive 10.0 m tall. If he comes to a stop 2.0 m below the surface of the water, what was his acceleration after he entered the water?
Answers
drwls
Vo = sqrt(2 a X)
Solve for acceleration, a.
X is the distance he goes below the water surface.
Vo is his speed when he hits the water.
Vo = sqrt(2 g H)
2 a X = 2 g H
a = g H / X
Solve for acceleration, a.
X is the distance he goes below the water surface.
Vo is his speed when he hits the water.
Vo = sqrt(2 g H)
2 a X = 2 g H
a = g H / X