A 0.500 kg is thrown vertically upward from the ground with an initial speed of 12.0 m/s. Define the initial potential energy to be zero on the ground. What are the potential energy and kinetic energy of the ball at 2.50m above the initial position?

To find the potential energy (PE) and kinetic energy (KE) of the ball at a height of 2.50m above the initial position, we need to consider the conservation of mechanical energy.

The total mechanical energy (E) of the system, which includes both potential and kinetic energy, remains constant throughout the motion. It can be expressed as:

E = PE + KE

Given that the initial potential energy is defined as zero, we can write the equation as:

E = 0 + KE

Since the ball is thrown vertically upward, it will experience a decrease in kinetic energy and an increase in potential energy as it reaches a height of 2.50m.

To calculate the final potential energy at this height, we can use the formula:

PE = m * g * h

where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (2.50m).

Given that the mass of the ball is 0.500 kg, we can substitute the known values into the formula:

PE = 0.500 kg * 9.8 m/s^2 * 2.50m

PE = 12.25 J

Therefore, the potential energy of the ball at 2.50m above the initial position is 12.25 Joules.

To find the kinetic energy at this height, we can subtract the potential energy from the total mechanical energy:

KE = E - PE

Since the total mechanical energy is conserved, it remains the same, and we know that the initial kinetic energy is given by the initial speed of the ball:

KE = E - PE = 0.500 kg * (12.0 m/s)^2 - 12.25 J

KE = 36 J - 12.25 J

KE = 23.75 J

Therefore, the kinetic energy of the ball at 2.50m above the initial position is 23.75 Joules.