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

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

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.

Well, it seems like you've done all the calculations correctly, but I have to say, that diver must really have a knack for breaking physics rules! I mean, falling upwards? That's impressive! Who knew water could have such a strong anti-gravity effect?

But let's bring you back to reality for a moment. In all seriousness, your calculations seem correct. The net force exerted by the water on the diver would indeed be 3.8E3 N (or 3800 N). So, it looks like the water is pushing the diver upwards with quite a significant force.

Keep up the good work, and who knows? Maybe one day you'll discover a new fundamental force of nature - the "water gravity-defying force"!

You are correct that the initial velocity for part b should be taken as the final velocity from part a, which is 14 m/s downwards. However, the equation f^2 = i^2 + 2ad should be modified to take into account the direction of the velocity and acceleration.

f^2 = i^2 - 2ad

Since the initial velocity is downwards and the deceleration is upwards, the negative sign is used. With this modification, the calculation becomes:

f^2 = (14 m/s)^2 - 2*(-9.80665 m/s^2)*(2.0 m)

f^2 = 196 m^2/s^2 + 39.2266 m^2/s^2

f^2 = 235.2266 m^2/s^2

Taking the square root of both sides gives:

f ≈ 15.34 m/s

So the final velocity when the diver comes to a stop 2.0 m below the surface of the water is approximately 15.34 m/s.

To find the net force exerted by the water, you correctly used the equation:

net force = water force - gravitational force

Substituting the given values:

net force = (65 kg)(-9.80665 m/s^2) - (65 kg)(15.34 m/s^2)

net force ≈ -637.9 N - 998.1 N

net force ≈ -1636 N

Therefore, the net force exerted by the water is approximately -1636 N, where the negative sign indicates it is acting in the opposite direction to the gravitational force.

To answer the first part of the question:

To find the diver's velocity when he hits the water, you can use the equation for the final velocity (vf) in vertical motion:

vf^2 = vi^2 + 2ad

where:
- vf is the final velocity (which is what we want to find)
- vi is the initial velocity (0 because the diver is initially not moving)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming downward as the negative direction)
- d is the displacement (10.0 m, because the diver jumped from a 10.0-m tower)

Plugging in these values, we get:

vf^2 = 0^2 + 2(-9.8)(-10.0)
vf^2 = 196
vf = √196
vf = 14 m/s (downward)

So you are correct in your calculation for the first part.

Now, for the second part:

To find the net force exerted by the water, we need to consider the acceleration of the diver as he comes to a stop in the water. In this case, the acceleration is in the upward direction because the diver is slowing down.

However, you were correct in your observation that the acceleration you calculated includes the acceleration due to gravity. To find the net force exerted by the water, we need to consider the acceleration due to gravity separately.

Taking the acceleration due to gravity as -9.8 m/s^2 (because it acts in the downward direction), we can find the net acceleration exerted by the water:

net acceleration = acceleration due to water - acceleration due to gravity
net acceleration = 49 m/s^2 - (-9.8 m/s^2)
net acceleration = 49 m/s^2 + 9.8 m/s^2
net acceleration = 59 m/s^2 (upward)

Now to find the net force exerted by the water, we can use Newton's second law:

net force = mass * net acceleration

Given that the mass of the diver is 65 kg, we have:

net force = 65 kg * 59 m/s^2
net force = 3,835 N (rounded to three significant figures)

So you were correct in your final calculation for the net force exerted by the water, which is approximately 3.8E3 N.