Suppose you throw a 0.058-kg ball with a speed of 11.0 m/s and at an angle of 31.5° above the horizontal from a building 12.3 m high.
(a) What will be its kinetic energy when it hits the ground?
(b) What will be its speed when it hits the ground?
the throwing angle has no effect
the KE at impact is the initial KE plus the KE gained from the gravitational potential (m g h)
KE = (.5 * .058 * 11.0^2) + (.058 * 9.8 * 12.3)
v^2 = (2 * KE) / .058
To solve this problem, we will use the principles of projectile motion and energy.
(a) To find the kinetic energy when the ball hits the ground, we need to calculate its initial potential energy at the top of the building and then convert it to kinetic energy.
The initial potential energy at the top of the building is given by the formula:
PE = m * g * h
Where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the building.
PE = 0.058 kg * 9.8 m/s^2 * 12.3 m
PE = 7.15 Joules
Since energy is conserved, the initial potential energy will be converted entirely into kinetic energy when the ball hits the ground.
KE = PE
KE = 7.15 Joules
Therefore, the kinetic energy of the ball when it hits the ground is 7.15 Joules.
(b) To find the speed of the ball when it hits the ground, we can use the principle of conservation of energy. The initial potential energy will be converted into the final kinetic energy.
The formula for kinetic energy is:
KE = (1/2) * m * v^2
Where m is the mass of the ball and v is its velocity or speed.
Setting the initial potential energy equal to the final kinetic energy:
KE = (1/2) * m * v^2
7.15 Joules = (1/2) * 0.058 kg * v^2
Simplifying the equation:
v^2 = (2 * KE) / m
v^2 = (2 * 7.15 Joules) / 0.058 kg
v^2 = 247.41 m^2/s^2
Taking the square root of both sides:
v = √(247.41 m^2/s^2)
v ≈ 15.72 m/s
Therefore, the speed of the ball when it hits the ground is approximately 15.72 m/s.