A spring loaded gun is used to shoot a 1.3 kg ball from a rooftop. When the gun is fired, a compressed spring will be released shooting the ball into the air. The ball will travel in a parabolic path eventually striking the ground below. The gun is at an angle of 39° above the horizontal. The spring has a spring constant of 1500 N/m and has been compressed by 0.8 m. The ball is 13 m above the ground when it leaves the gun. Air resistance is not present.
A) How much Elastic Energy does the gun have before being fired?
B) How much Kinetic Energy does the ball have as it leaves the gun?
C) What is the ball's velocity as it leaves the gun??
D) How much Gravitational Energy does the ball have as it leaves the gun?
E) How much Kinetic Energy does the ball have at its highest point? HINT: If you know the balls velocity as it leaves the gun you should easily be able to figure out the ball's velocity at its highest point.
F) How much Gravitational Energy does the ball have at its highest point?
G) How high is the ball above the ground at its highest point?
H) How much Kinetic Energy does the ball have as it hits the ground?
I) How much Gravitational Energy does the ball have as it hits the ground?
Honors Physics
A) To calculate the elastic energy of the gun before being fired, we can use the formula for elastic potential energy:
Elastic Energy = (1/2) * k * x^2
Where k is the spring constant (1500 N/m) and x is the compression distance (0.8 m).
Elastic Energy = (1/2) * 1500 * 0.8^2
B) To calculate the kinetic energy of the ball as it leaves the gun, we can use the formula for kinetic energy:
Kinetic Energy = (1/2) * m * v^2
Where m is the mass of the ball (1.3 kg) and v is the velocity of the ball.
C) The velocity of the ball as it leaves the gun can be calculated using the projectile motion equations. We can start by finding the initial velocity in the x-direction and y-direction.
v0x = v0 * cosθ
Where θ is the angle of the gun above the horizontal (39°) and v0 is the initial velocity of the ball.
v0y = v0 * sinθ
Since the ball is launched horizontally in the x-direction, v0x = v0 * cosθ = 0.
Therefore, v0 * sinθ = v0y
Now, we can calculate the magnitude of the initial velocity (v0) using the vertical motion equation:
y = v0y * t - (1/2) * g * t^2
Where y is the height of the ball above the ground when it leaves the gun (13 m) and g is the acceleration due to gravity (9.8 m/s^2).
13 = v0 * sinθ * t - (1/2) * 9.8 * t^2
We can solve this equation for t:
t = (v0 * sinθ + sqrt((v0 * sinθ)^2 + 2 * 9.8 * 13)) / 9.8
Substituting this value of t back into the equation for v0y, we can solve for v0:
v0 = (y + (1/2) * g * t^2) / t
D) The gravitational potential energy of the ball as it leaves the gun can be calculated using the formula:
Gravitational Energy = m * g * h
Where m is the mass of the ball (1.3 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ball above the ground when it leaves the gun (13 m).
Gravitational Energy = 1.3 * 9.8 * 13
E) At the highest point of the ball's trajectory, its velocity in the y-direction will be 0. We can use this information to find the velocity at the highest point. By symmetry, the velocity in the y-direction when the ball reaches the highest point will be equal in magnitude but opposite in direction to the initial velocity in the y-direction.
Therefore, the velocity at the highest point is equal to v0y.
F) The gravitational potential energy of the ball at its highest point can be calculated using the formula mentioned earlier:
Gravitational Energy = m * g * h
Where m is the mass of the ball (1.3 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ball above the ground at its highest point.
G) To find the height of the ball above the ground at its highest point, we need to determine the time it takes for the ball to reach the highest point.
Using the equation for vertical motion:
v = v0 - g * t
At the highest point, the velocity in the y-direction is 0. Therefore:
0 = v0y - g * t
We can solve this equation for t:
t = v0y / g
Now, we can use this value of t to find the height at the highest point:
h = y + v0y * t - (1/2) * g * t^2
H) To calculate the kinetic energy of the ball as it hits the ground, we use the same formula as in part B:
Kinetic Energy = (1/2) * m * v^2
I) The gravitational potential energy of the ball as it hits the ground can be calculated using the formula mentioned earlier:
Gravitational Energy = m * g * h
Where m is the mass of the ball (1.3 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground when it hits the ground.