A 45.0 kg diver steps off a 8.0 m high diving board and drops straight down into the water. If the diver comes to rest 5.0 m below the surface of the water, determine the average resistance force exerted on the diver by the water.

Answer 1

well, there is a rick you can use here to make it easy. If the force is constant, then the average speed is halfway between the starting speed and the final speed (which is zero here.
So what is the speed when he hits the water?
v final = sqrt (2 g h) =sqrt(2*9.81 * 8)
= 12.5 m/s
NOW the slow down
net force up = m a = rate of change of momentum
I will assume neutral buoyancy so weight down = displacement force up
then
net force up = water drag
how long to stop?
average speed in water = 6.26 m/s
so time to stop = 5m/6.26m/s = .798 second
change in momentum = 45 * 12.5 = 562.5 kg m/s
(NOTE - this ignores the "added mass" of the water. Water around the diver is accelerated with the diver and thus requires more force in real life)
so in the end
Faverage = 562.5 / .798 = 705 Newtons

Answer 2

another way (easier)

The force F times 5 meters = potential energy lost in water
= m g (height above water)
note no pe change below water because buoyancy is about the same as weight
5 F = 45 * 9.81 * 8
F = 706 Newtons