A cliff diver,mass 75kg,free falls off a cliff that is 50m above the water surface.He hits the water surface and stops 5m under the surface.calculate:

4.1 The net force while free falling.
4.2 The divers' velocity when he reaches the water surface.
4.3 calculate the time it takes the water to stop the diver.

net force when falling= zero

velocity at water hit: 1/2 m v^2=mg*50
v= sqrt(100g)

time to stop in water is rather complicated, unless you assume the acceleration is constant at hitting until stopping (a ridiculous assumption), and that force is constant (another ridiculous assumption).

With those assumptions
Vf^2=Vi^2+2ad where a= force/mass
solve for force

To answer these questions, we will need to use the principles of physics. Let's break it down step by step:

4.1 The net force while free falling:
When the cliff diver is free falling, the only force acting on them is gravity. According to Newton's second law of motion, the net force acting on an object is equal to its mass multiplied by its acceleration. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. So, the net force acting on the diver while free falling can be calculated as follows:

Net force = mass x acceleration
Net force = 75 kg x 9.8 m/s^2
Net force = 735 N

Therefore, the net force while free falling is 735 Newtons.

4.2 The diver's velocity when he reaches the water surface:
When the diver reaches the water surface, he is no longer free falling but decelerating due to the resistance of the water. To calculate the velocity, we need to use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity (which is zero when the diver starts free falling)
a = acceleration (which is the acceleration due to gravity)
s = displacement (the distance fallen, which is 50 m)

Rearranging the equation, we get:

v = sqrt(2as)

v = sqrt(2 x 9.8 m/s^2 x 50 m)
v ≈ 31.3 m/s

Therefore, the diver's velocity when he reaches the water surface is approximately 31.3 m/s.

4.3 The time it takes the water to stop the diver:
To calculate the time it takes for the water to stop the diver, we need to use the equation of motion:

v = u + at

Where:
v = final velocity (which is zero when the diver stops)
u = initial velocity (which is 31.3 m/s, as calculated above)
t = time taken
a = acceleration (which is the deceleration due to the water's resistance)

Rearranging the equation, we get:

t = (v - u) / a

t = (0 - 31.3 m/s) / a

Now, the deceleration due to the water's resistance depends on various factors like the diver's shape, the angle of entry, and the density of the water. Without specific information, it is difficult to provide an accurate value. However, the deceleration due to water resistance is generally quite significant, so we can estimate it to be around 30 m/s^2.

t ≈ (0 - 31.3 m/s) / -30 m/s^2
t ≈ 1.04 s

Therefore, it would take approximately 1.04 seconds for the water to stop the diver.