John Stands at the edge of a flat roof of a three story building and throws a 180g ball straight up into the air that reaches a height of 3.0 meters above the roof. ( building is 14 meters high)

A. What is the initial speed of the ball?
B. How long does it take to reach it's maximum height?
C. Assuming it misses the building on it's way down, at what speed does it hit the ground? ( building is 14 meters high)
D. What is the total length of the time the ball is in the air?
E. What is the intitial momentum of the ball?
F. What is the momentum of the ball at its peak?
G. What is the momentum of the ball when it hits the ground?

Thank you!

a. Vf^2 = Vi^2 + 2*a*d

solve for Vi given Vf = 0 (speed at max height is 0), a = -9.81 m/s^2 and d = 3m.

b. Once you know Vi, you can use
Vf = a*t + Vi and solve for t.
(once again, Vf = 0 since you're only considering the path of the ball from the top of the building to maximum height)

c. Vf^2 = Vi^2 + 2*a*d
(d = height of building + 3m)

d. Vf = a*t + Vi
Use your answer from part c. to solve for the time it takes the ball to fall from max height to the ground and then add your answer from part b. to get the total time.

e,f,g: momentum = m*v

To solve these physics problems, we can use the equations of motion. Here's how we can find the answers:

A. To find the initial speed of the ball, we need to know the final speed when it reaches the maximum height. Since the ball has zero speed at its maximum height, the initial speed is equal to the final speed. We know the ball goes up by 3.0 meters, and we can use the kinematic equation:

vf^2 = vi^2 + 2aΔx

where vf is the final velocity (0 m/s), vi is the initial velocity (what we need to find), a is the acceleration due to gravity (-9.8 m/s^2), and Δx is the displacement (3.0 m). Rearranging the equation, we get:

vi = √(vf^2 - 2aΔx)
= √(0 - 2*(-9.8)*3.0)
≈ 7.67 m/s

So the initial speed of the ball is approximately 7.67 m/s.

B. The time taken to reach the maximum height can be found using the equation:

vf = vi + at

where vf is the final velocity (0 m/s at maximum height), vi is the initial velocity (7.67 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken (what we need to find). Rearranging the equation, we get:

t = (vf - vi) / a
= (0 - 7.67) / (-9.8)
≈ 0.78 s

So it takes approximately 0.78 seconds to reach the maximum height.

C. Assuming the ball misses the building on its way down, it will fall freely under gravity. The speed of the ball when it hits the ground can be found using the equation:

vf^2 = vi^2 + 2aΔx

where vf is the final velocity (what we need to find), vi is the initial velocity at maximum height (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and Δx is the displacement (14 m). Rearranging the equation, we get:

vf = √(vi^2 + 2aΔx)
≈ √(0 + 2*(-9.8)*14)
≈ 19.54 m/s

So the ball hits the ground with approximately 19.54 m/s.

D. The total time the ball is in the air can be found by multiplying the time taken to reach the maximum height by 2, as the ball takes the same amount of time to fall back down. Therefore, the total time is approximately:

2 * 0.78 s = 1.56 s

E. The initial momentum of the ball can be calculated using the equation:

momentum = mass * velocity

The mass of the ball is given as 180 grams (0.18 kg), and the initial velocity is 7.67 m/s. Therefore, the initial momentum is:

momentum = 0.18 kg * 7.67 m/s
≈ 1.38 kg·m/s

F. At the peak of its trajectory, the ball momentarily stops, so its momentum is zero.

G. The momentum of the ball when it hits the ground can be calculated using the equation:

momentum = mass * velocity

The mass of the ball is still 0.18 kg, and the velocity is calculated earlier as 19.54 m/s. Therefore, the momentum is:

momentum = 0.18 kg * 19.54 m/s
≈ 3.52 kg·m/s

So the momentum of the ball when it hits the ground is approximately 3.52 kg·m/s.