a 0.15 kg ball is thrown into the air & rises to a height of 20.0 m. how much kinetic energy did the ball initially have? ? ?

Vo^2 = Vf^2 - 2g*d.

Vo^2 = 0 -(-19.6)*20 = 392.
Vo = 19.8 m/s.
KE = 0.5m*V^2 = 0.075 * (19.8)^2 = 29.4
Joules.

thankss . what does the ^ mean tho ?

to the power of...

To determine the initial kinetic energy of the ball, we need to use the formula for kinetic energy, which is given by:

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

Since we are not provided with the velocity of the ball, we can find it using the concept of conservation of energy. The total mechanical energy of the ball is conserved throughout its motion, meaning that the sum of its kinetic energy (KE) and potential energy (PE) remains constant.

At the highest point (when the ball reaches a height of 20.0 m), all of its initial kinetic energy is converted into potential energy. Therefore, we can set up the equation as follows:

KE_initial + PE_initial = KE_final + PE_final

Since the ball starts from the ground, its initial potential energy (PE_initial) is zero. At its highest point, the final potential energy (PE_final) will be equal to the initial potential energy because no additional work is done on the ball. Therefore, we have:

KE_initial + 0 = 0 + PE_final

Since the ball reaches a height of 20.0 m, we can calculate the potential energy (PE_final) using the formula:

Potential Energy = mass * gravitational acceleration * height

Given:
mass = 0.15 kg
height = 20.0 m
gravitational acceleration (g) = 9.8 m/s^2

PE_final = 0.15 kg * 9.8 m/s^2 * 20.0 m = 29.4 J

Now, substituting this back into the conservation of energy equation, we have:

KE_initial + 0 = 0 + 29.4 J

Therefore,

KE_initial = 29.4 J

Hence, the initial kinetic energy of the ball is 29.4 Joules.