A 65-kg diver jumps off a 10.0-m tower.

Question

a) Find the diver's velocity when he hits the water.
b) The diver comes to a stop 2.0m below the surface.
Find the net force exerted by the water.

I get the first part: sqrt(2ad), thus about 14 m/s downwards. For part b, I have...
f = final velocity
i = initial velocity
a = acceleration
d = displacement

f^2 = i^2 + 2ad
0 = 14^2 + 2*a*2.0
-(14^2) = 4.0*a
a = -(14^2)/4.0
a = -49 down, thus 49 m/s^2 upwards

What I'm confused about is doesn't this take into account gravity already? Thus isn't the average acceleration from the water 59 m/s^2 upwards?

net = water - gravity
49 = x - 9.80665
x = 59

If this is true, this would thus mean that part b would be 3.8E3 N (I think).

F=ma
F=65*59=3.8E3 N

Answer 1

The acceleration should be separated not averaged. The correct answer for part be is 2548N and this is how:

Fnet= F2+F1
in order to find F1 and F2, use F=ma.

The acceleration for F1 is gravity as it's the part where the diver hits the water. Mass is 65kg.
F1= 65kg*9.8m/s^2= 637N
The acceleration for F2 is the part where the diver goes from the 0m to 2m below and is found through the kinematic equation of Vf^2=Vi^2+2ad
a= deceleration of 49m/s^2.Mass is still 65kg.
F2= 65kg*-49m/s^2=-3185N

Now that you have both F1 and F2, add.
Fnet= 637N+(-3185N)= -2548N.
This final answer is the net force of the diver so according to Newton's Third Law, the force exerted by the water is equal and opposite of the diver's.