A 50 kg diver leaps off the 20 meter diving board with a velocity of 5 m/s. What is his kinetic energy just before hitting the water?

What is direction of initial velocity?

If v=5 m/s is downward velocity,
PE+KE1=KE2
mgh+mv1²/2 =mv2²/2
v2 =sqrt(2gh+v1²)

F=ma

F=50(5)= 250

M(k)=F(k)/F(n)
M(k)=250/(50(9.8))= .51 is the coefficent of his kinetic energy while 250 is his kinetic force.

Probably wrong but that's how I would have done it

To calculate the kinetic energy of the diver just before hitting the water, we can use the formula:

Kinetic Energy = 1/2 * mass * velocity^2

Given:
Mass (m) = 50 kg
Velocity (v) = 5 m/s

Plugging in the values:

Kinetic Energy = 1/2 * 50 kg * (5 m/s)^2

Simplifying the equation:

Kinetic Energy = 1/2 * 50 kg * 25 m^2/s^2

Kinetic Energy = 1250 kg·m^2/s^2

Therefore, the kinetic energy of the diver just before hitting the water is 1250 kg·m^2/s^2.

To calculate the kinetic energy of the diver just before hitting the water, we need to use the formula for kinetic energy, which is given by:

Kinetic Energy = (1/2) * mass * velocity^2

Given:
Mass of the diver (m) = 50 kg
Velocity of the diver (v) = 5 m/s

Using the values given, we can now substitute them into the formula to calculate the kinetic energy of the diver:

Kinetic Energy = (1/2) * 50 kg * (5 m/s)^2

First, we square the velocity:
Velocity^2 = 5 m/s * 5 m/s = 25 m^2/s^2

Now, substituting the values into the formula:
Kinetic Energy = (1/2) * 50 kg * 25 m^2/s^2

Next, we calculate the product of 50 kg and 25 m^2/s^2:
Kinetic Energy = 0.5 * 50 kg * 25 m^2/s^2 = 625 Joules

Therefore, the kinetic energy of the diver just before hitting the water is 625 Joules.