Compressed air is used to fire a 50g ball vertically upward from a 0.60-m-tall tube. The air exerts an upward force of 3.0N on the ball as long as it is in the tube.

How high does the ball go above the top of the tube?

work input=force*distance=3*.6 Joules

PE at top: mgh
mgh=1.8J
solve for height h.

some PE is gainied from work going up the tube, so h above yields the height it went above the bottom of the tube, so how high above the top of the tube, subract .6m

6.89m

To calculate the height the ball reaches above the top of the tube, we can use the principle of conservation of energy. The initial potential energy of the ball inside the tube is equal to the work done by the air compressor.

The potential energy (PE) of an object at a certain height is given by the formula PE = mgh, where m is the mass of the object (in kg), g is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and h is the height (in meters).

Given:
Mass of the ball, m = 50 g = 0.05 kg
Height of the tube, h = 0.60 m
Force exerted by the air, F = 3.0 N

Since the force exerted by the air is constant and in the opposite direction to the displacement, the work done is given by the formula W = -FΔh, where Δh is the change in height.

W = -FΔh
Work done by the air, W = -3.0 N × 0.60 m = -1.8 J (since the displacement is in the opposite direction of the force)

According to the conservation of energy, the initial potential energy inside the tube is equal to the final potential energy at the highest point of motion. Therefore,

Initial PE + Work done = Final PE
mgh + (-1.8 J) = 0 + mgh'

Simplifying the equation gives us:
mgh' = 1.8 J

Substituting the given values:
(0.05 kg)(9.8 m/s²)h' = 1.8 J

Solving for h', we find:
h' = 1.8 J / (0.05 kg)(9.8 m/s²)
h' ≈ 3.67 m

Therefore, the ball reaches a height of approximately 3.67 meters above the top of the tube.