A 0.20kg ball is thrown straight up from the edge of a 30.0m tall building at a velocity of 22.0m/s. The
ball moves up to the maximum height and then falls to the ground at the base of the building. Use the
law of conservation of energy to answer the following questions, assuming that the reference level for
gravitational potential energy is ground level.
a) What is the total energy of the ball at the start when it had a velocity of 22m/s?
b) What is the velocity of the ball at the maximum height? What is the maximum height of the ball?
c) What is the velocity of the ball when it hits the ground?
a) The total energy of the ball at the start, when it had a velocity of 22 m/s, can be calculated using the equation:
Total energy = kinetic energy + potential energy
Kinetic energy = 1/2 * mass * velocity^2
Potential energy = mass * gravity * height
Given:
Mass of the ball = 0.20 kg
Velocity = 22 m/s
Height = 30.0 m
Gravity = 9.8 m/s^2
Kinetic energy = 1/2 * 0.20 kg * (22 m/s)^2 = 48.4 J
Potential energy = 0.20 kg * 9.8 m/s^2 * 30.0 m = 58.8 J
Total energy = 48.4 J + 58.8 J = 107.2 J
Therefore, the total energy of the ball at the start is 107.2 Joules.
b) At the maximum height, all the initial kinetic energy is converted into potential energy. So, the total energy at the maximum height remains the same as at the start, which is 107.2 J.
At maximum height, potential energy is equal to the total energy:
Potential energy = mass * gravity * height
58.8 J = 0.20 kg * 9.8 m/s^2 * height
Height = 58.8 J / (0.20 kg * 9.8 m/s^2) = 30.0 m
Therefore, the maximum height of the ball is 30.0 meters.
To find the velocity of the ball at the maximum height, we can use the equation:
Total energy = kinetic energy + potential energy
Since the total energy is the same at the start and at the maximum height, we can rearrange the equation:
kinetic energy = Total energy - potential energy
kinetic energy = 107.2 J - 58.8 J = 48.4 J
Now we can use the equation for kinetic energy to find the velocity at the maximum height:
Kinetic energy = 1/2 * mass * velocity^2
48.4 J = 1/2 * 0.20 kg * velocity^2
48.4 J = 0.1 kg * velocity^2
velocity^2 = 484 J / 0.1 kg
velocity^2 = 4840 m^2/s^2
velocity = √(4840 m^2/s^2)
velocity = 69.7 m/s (rounded to 3 decimal places)
Therefore, the velocity of the ball at the maximum height is 69.7 m/s.
c) When the ball hits the ground, the potential energy is zero since the ball is at ground level. So, all the energy is in the form of kinetic energy.
Using the equation for kinetic energy:
Kinetic energy = 1/2 * mass * velocity^2
0.20 kg * velocity^2 = 107.2 J
velocity^2 = 107.2 J / 0.20 kg
velocity^2 = 536 m^2/s^2
velocity = √(536 m^2/s^2)
velocity = 23.1 m/s (rounded to 3 decimal places)
Therefore, the velocity of the ball when it hits the ground is 23.1 m/s.