Chapter 6: Q. 44 (page 156)

Compressed air is used to fire a 50 g ball vertically upward from a 1.0 m-tall tube. The air exerts an upward force of 2N on the ball as long as it is in the tube. How high does the ball go above the top of the tube?

Short Answer

Expert verified

The ball go 3.1 m above the top of the tube.

Step by step solution

Step 1. Find the acceleration of the ball

Knowing the mass of the ball and the force acting on it.

So, projecting the forces on the vertical axis and setting the positive direction upward, we have:

$F - m g = m a \Rightarrow a = \frac{F - m g}{m}$

The speed at the top of the tube, knowing that there was no initial velocity

$\begin{array}{r} v^{2} = v_{0}^{2} + 2 a h \\ \Rightarrow v = \sqrt{2 a h} \\ \Rightarrow \sqrt{\frac{2 (F - m g) h}{m}} = \sqrt{\frac{2 \times (2 N - (0.05 k g \times 9.8 m / s^{2})) \times 1 m}{0.05 k g}} \\ = 7.78 m / s \end{array}$

$= 7.78$ m/s

Step 2. Calculate the rise time

Knowing the vertical speed when the ball exits the tube, we can first calculate the rise time,

$v_{f} = v_{i} + g t_{r} \Rightarrow t_{r} = - \frac{v_{i}}{g} \Rightarrow t_{r} = - \frac{7.78 m / s}{- 9.8 m / s^{2}}$

$= 0.794$ s

Step 3. Find the maximum height

Now substitute this time in the second equation of motion to find the maximum height. From the first equation of motion, setting the positive direction upward,

$H = u_{1} t + \frac{g {t_{r}}^{2}}{2} = (7.78 m / s) \cdot (0.794 s) + \frac{(- 9.8 m / s^{2}) \cdot {(0.794 s)}^{2}}{2}$