a ball (mass = 0.250 kg) is kicked off a porch (h = 1.50 m above the ground) with an initial velocity of 17.0 m/s at an unknown angle. what is the initial KE. What is the initial PE. what is the KE and speed of the ball when it is at the top of its ballistic arc (h = 9.50 m). What is the speed of the ball when it reaches the ground - (h = 0).
To find the initial kinetic energy (KE) of the ball, we can use the equation:
KE = (1/2) * m * v^2
where m is the mass of the ball and v is the magnitude of its initial velocity.
Plugging in the given values:
m = 0.250 kg
v = 17.0 m/s
KE = (1/2) * 0.250 kg * (17.0 m/s)^2
KE = 36.125 J
So, the initial kinetic energy of the ball is 36.125 Joules.
To find the initial potential energy (PE), we can use the equation:
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 initial height above ground.
Plugging in the given values:
m = 0.250 kg
h = 1.50 m
PE = 0.250 kg * 9.8 m/s^2 * 1.50 m
PE = 3.675 J
So, the initial potential energy of the ball is 3.675 Joules.
At the top of its ballistic arc, the ball reaches a height of 9.50 m. To find its kinetic energy and speed at this point, we need to determine its velocity (v) using the energy conservation principle.
At the top of its arc, all of the initial potential energy is converted to kinetic energy (no energy losses due to friction or air resistance). Therefore, we can equate the initial potential energy to the kinetic energy at the top:
PE = KE
Using the equation for potential energy:
m * g * h = (1/2) * m * v^2
Simplifying:
g * h = (1/2) * v^2
Plugging in the given values:
g ≈ 9.8 m/s^2
h = 9.50 m
9.8 m/s^2 * 9.50 m = (1/2) * v^2
v^2 = (9.8 m/s^2 * 9.50 m) / (1/2)
v^2 = 93.41 m^2/s^2
v ≈ 9.67 m/s
So, the velocity (and speed) of the ball at the top of its ballistic arc is approximately 9.67 m/s.
To find the speed of the ball when it reaches the ground (h = 0), we can use the conservation of mechanical energy again since there are no energy losses. The initial potential energy equals the final kinetic energy:
PE = KE
m * g * h = (1/2) * m * v^2
Plugging in the given values:
g ≈ 9.8 m/s^2
h = 9.50 m
9.8 m/s^2 * 9.50 m = (1/2) * v^2
v^2 = (9.8 m/s^2 * 9.50 m) / (1/2)
v^2 = 93.41 m^2/s^2
v ≈ 9.67 m/s
So, the velocity (and speed) of the ball at the top of its ballistic arc is approximately 9.67 m/s.
When the ball reaches the ground (h = 0), it loses its potential energy and gains the same amount of kinetic energy. Thus, its kinetic energy at this point will be the initial potential energy:
KE = PE = 3.675 J
Therefore, the speed of the ball when it reaches the ground is determined by the equation:
KE = (1/2) * m * v^2
Plugging in the given values:
m = 0.250 kg
KE = 3.675 J
3.675 J = (1/2) * 0.250 kg * v^2
v^2 = (3.675 J * 2) / 0.250 kg
v^2 = 29.4 m^2/s^2
v ≈ 5.42 m/s
So, the speed of the ball when it reaches the ground is approximately 5.42 m/s.