A high diver of mass 68.2 kg jumps off a board 10.4 m above the water.

The acceleration of gravity is 9.8 m/s2 .
If his downward motion is stopped 3.58 s after he enters the water, what average upward force did the water exert on him?
I already calculated that his velocity before hitting the water is 14.277 m/s.

mgh=mv₀²/2

v₀=sqrt(2gh)=sqrt(2•9.8•10.4)=14.28 m/s
v= v₀-at, v=0 =>
a= v₀/t = 14.28/3.58 =4 m/s²
F=ma=68.2•4=272.8 N

To find the average upward force exerted by the water on the diver, we can use the concept of impulse. Impulse is given by the product of force and the time interval over which it acts.

Since the diver's downward motion is stopped by the water, the average upward force exerted on the diver by the water is equal to the change in momentum of the diver.

We can find the initial momentum of the diver using the equation:
Initial Momentum = mass × velocity

Initial momentum = 68.2 kg × 14.277 m/s = 974.9034 kg⋅m/s

The final momentum of the diver is zero since his downward motion is stopped.

Now, the change in momentum is calculated as:
Change in momentum = Final momentum - Initial momentum
Change in momentum = 0 - 974.9034 kg⋅m/s

The change in momentum represents the impulse, which is equal to the average upward force (since the time interval is given).

Impulse = Average upward force × time

Substituting the known values:
Change in momentum = Average upward force × 3.58 s

0 - 974.9034 kg⋅m/s = Average upward force × 3.58 s

Solving for Average upward force:
Average upward force = (0 - 974.9034 kg⋅m/s) / 3.58 s

Average upward force ≈ -271.96 N

Since the upward force is given a negative sign, it indicates that the force is in the opposite direction of the downward motion. Therefore, the average upward force exerted by the water on the diver is approximately 271.96 N, acting in the upward direction.

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 its mass multiplied by its acceleration. In this case, we need to find the net force acting on the diver while he is in the water.

First, we can calculate the acceleration experienced by the diver while he is in the water. We know that his initial velocity before hitting the water is 14.277 m/s, and his final velocity is 0 m/s (since his downward motion is stopped). The time taken for his downward motion to be stopped is 3.58 s.

We can use the equation of motion:
v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. By rearranging the equation, we can solve for the acceleration:

a = (v - u) / t

Substituting the given values, we get:
a = (0 - 14.277) / 3.58

a ≈ -3.987 m/s^2 (Note: the negative sign indicates that the motion is in the opposite direction to the initial velocity)

Now, we can find the net force acting on the diver while he is in the water using Newton's second law:

Fnet = m * a

Substituting the given mass of the diver (m = 68.2 kg) and the calculated acceleration (a = -3.987 m/s^2), we get:

Fnet = 68.2 * (-3.987)

Fnet ≈ -271.88 N

Since we are interested in the average upward force exerted by the water, we take the absolute value of the net force:

Average upward force = |Fnet| ≈ 271.88 N

Therefore, the average upward force exerted by the water on the diver is approximately 271.88 Newtons.