a 5.7 kg ball rolls over the edge of a 4.4 m tall cliff. If the total mechanical energy of the ball is 890 j at the instant that it goes over the edge, how fast is the ball moving?

Not sure how to get V when you know TME.

The answer depends upon whether the ball is hollow inside, like a basketball or tennis ball, or solid and of uniform density.

Total kinetic energy is
(1/2)MV^2 + (1/2) I w^2
in either case

where w is the angular rotation rate, V/R, and I is the moment of inertia

That just confused me more cause it's just asking for Velocity and we haven't talked anything about w or i

To determine the speed (V) of the ball when it goes over the edge of the cliff, you can use the principle of conservation of mechanical energy. At the edge of the cliff, the potential energy (PE) of the ball is converted into kinetic energy (KE) since it is now moving.

First, you need to calculate the potential energy of the ball before it falls using the formula:

PE = m * g * h

Where:
m is the mass of the ball (5.7 kg)
g is the acceleration due to gravity (9.8 m/s^2)
h is the height of the cliff (4.4 m)

PE = 5.7 kg * 9.8 m/s^2 * 4.4 m
PE = 244.584 J

Since the total mechanical energy (TME) is given as 890 J, the difference between the initial total mechanical energy and potential energy is equal to the kinetic energy of the ball at the edge of the cliff.

KE = TME - PE
KE = 890 J - 244.584 J
KE = 645.416 J

Now, you can use the formula for kinetic energy to find the speed (V) of the ball:

KE = 0.5 * m * V^2

Substituting the values:

645.416 J = 0.5 * 5.7 kg * V^2

Now solve for V:

V^2 = (645.416 J) / (0.5 * 5.7 kg)
V^2 = 225.198 m^2/s^2

Finally, take the square root of both sides to get the speed (V):

V = √(225.198 m^2/s^2)
V ≈ 15.01 m/s

Therefore, the ball is moving at approximately 15.01 m/s when it goes over the edge of the cliff.