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 can be calculated using the formula:
Total energy = Kinetic energy + Potential energy
Given:
Mass of the ball (m) = 0.20 kg
Velocity of the ball (v) = 22 m/s
The kinetic energy of the ball is given by:
Kinetic energy = (1/2) * mass * velocity^2
Substituting the given values:
Kinetic energy = (1/2) * 0.20 kg * (22 m/s)^2
Kinetic energy = 0.20 kg * 242 m^2/s^2
Kinetic energy = 48.4 J (Joules)
The potential energy of the ball is given by:
Potential energy = mass * g * height
Where g is the acceleration due to gravity and equals 9.8 m/s^2, and height is 30.0 m.
Potential energy = 0.20 kg * 9.8 m/s^2 * 30.0 m
Potential energy = 58.8 J (Joules)
Therefore, the total energy of the ball at the start is:
Total energy = Kinetic energy + Potential energy
Total energy = 48.4 J + 58.8 J
Total energy = 107.2 J (Joules)
b) At the maximum height, the ball has reached its potential energy and its kinetic energy is zero. Therefore, the total energy is equal to the potential energy.
Given:
Potential energy = 58.8 J (from part a))
To find the maximum height, we can rearrange the potential energy formula:
Potential energy = mass * g * height
So, height = Potential energy / (mass * g)
height = 58.8 J / (0.20 kg * 9.8 m/s^2)
height = 30.0 m
Therefore, the maximum height of the ball is 30.0 m.
Since the ball has reached its maximum height, its velocity at this point is zero.
c) To find the velocity of the ball when it hits the ground, we can use the principle of conservation of energy. The total energy of the ball when it hits the ground is equal to the potential energy at the maximum height.
Given:
Potential energy = 58.8 J (from part a))
To find the velocity, we can use the following formula:
Potential energy = (1/2) * mass * velocity^2
58.8 J = (1/2) * 0.20 kg * velocity^2
Solving for velocity:
velocity^2 = (2 * 58.8 J) / 0.20 kg
velocity^2 = 588 J / 0.20 kg
velocity^2 = 2940 m^2/s^2
velocity = sqrt(2940 m^2/s^2)
velocity ≈ 54.09 m/s
Therefore, the velocity of the ball when it hits the ground is approximately 54.09 m/s.