A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 1.80 s after her feet first touch the water, what average upward force did the water exert on her?

Fav*(1.80 s) = Momentum when entering water

To find the average upward force exerted by the water on the high diver, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, we need to find the acceleration of the high diver.

First, let's find the time it takes for the diver to fall from the board to the water. We can use the equation of motion:

h = (1/2)gt²

where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation gives us:

t = √(2h/g)

Plugging in the values given, we have:

t = √(2 * 10.0 m / 9.8 m/s²)
t = √(20.0 / 9.8) s
t ≈ 2.02 s

This tells us that it takes approximately 2.02 seconds for the high diver to fall from the board to the water.

Next, we can calculate the acceleration using the following equation:

a = (vf - vi) / t

where vf is the final velocity, vi is the initial velocity (which is 0 in this case since the diver starts from rest), and t is the time. Rearranging the equation gives us:

a = vf / t

Now, the final velocity of the diver can be calculated using the equation of motion:

vf = vi + at

Since the diver starts from rest, vi = 0, so we have:

vf = at

Plugging in the values, we have:

vf = a * t
vf = a * 2.02 s

Now, we know that the final velocity of the diver when her downward motion is stopped is 0 m/s.

Therefore, we can set vf = 0 and solve for the acceleration (a):

0 m/s = a * 2.02 s
a ≈ 0 m/s²

Since the acceleration is approximately 0 m/s², it means that the diver experienced zero net force during her fall. This is because her downward force due to gravity is balanced by the equal and opposite upward force exerted by the water. Therefore, the average upward force exerted by the water on the diver is equal to her weight:

F = mg

F = (60.0 kg) * (9.8 m/s²)

F ≈ 588 N

Hence, the average upward force exerted by the water on the high diver is approximately 588 N.