Question

Compressed air is used to fire a 46 g ball vertically upward from a 0.60-m -tall tube. The air exerts an upward force of 2.8 N on the ball as long as it is in the tube.

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

Answer 1

Mass m = 46 g = 0.046 kg
When inside the tube: -
Force by compressed air = 2.8 N upward
Force by gravity = mg = 0.046 * 9.8 = 0.4508 N downward
Therefore, net force F = 2.8 N - 0.4508 N = 2.34 N upward
Acceleration a = F/m = 2.34/0.046 = 50.86 m/s^2 upward
Taking upward direction as positive,
velocity at the bottom of the tube = u = 0
Displacement S = 1.0 m
Acceleration a = 50.86 m/s^2
Let v1 = velocity at the top of the tube
v1^2 = u^2 + 2aS
Or v1^2 = 0 + 2.8 * 50.86 * 0.60
Or v1^2 = 85.44
Or v1 = sqrt(85.44)
Or v1 = 9.24m/s

After the ball comes out of the tube, its acceleration is g downward.
Therefore, the height reached = v1^2/(2g)
= 9.24^2/(2.8 * 9.8)
= 3.11 m