A high diver of mass 50.0 kg jumps off a board 10.0 m above the water. If her downward motion is stopped 1.80 s after she enters the water, what average force did the water exert on her?

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

First, let's determine the acceleration of the diver when she enters the water. We can use the kinematic equation:

d = vit + (1/2)at^2

Where:
- d is the distance covered (10.0 m in this case)
- vi is the initial velocity (which is 0 since the diver starts from rest)
- a is the acceleration (which we need to find)
- t is the time taken (1.80 s in this case)

Rearranging the equation to solve for acceleration:

a = 2(d - vit) / t^2

Substituting the values:

a = 2(10.0 - 0) / (1.80)^2
a = 2 * 10.0 / 3.24
a ≈ 6.17 m/s^2

Now that we have the acceleration, we can use Newton's second law to find the average force:

F = ma

Substituting the values:

F = (50.0 kg) * (6.17 m/s^2)
F ≈ 308.5 N

Therefore, the average force exerted by the water on the diver is approximately 308.5 Newtons.