A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he enters the water, what average upward force did the water exert on him

1120 N

Explanation:

The velocity with which he hits the water can be found with kinematics:

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

F = 1120 N