Answer to Question #90822 in Mechanics | Relativity for Paul

Question

A hot-air balloon is ascending at the rate of 14 m/s at a height of 95 m above the ground when a package is dropped. (a) How long does the package take to reach the ground? (b) What is the speed of the package?

Answer 1

To approach the problem, divide it into two parts: the initial conditions and the dropped package.

First, we know that the balloon is ascending with the speed of 14 m/s. This will be the initial speed of the package - it is 14 m/s directed upward. The height at which the package starts its decelerating motion upward is 95 m, then it stops and starts falling.

So, calculate the time required for the package to ascend and stop:

$$t_\text{up}=\frac{v}{g}=\frac{14}{9.8}=1.43\text{ s}.$$

This height above the 95 meters is

$$h_\text{up}=\frac{v^2}{2g}=\frac{14^2}{2\cdot9.8}=10 \text{ m}.$$

So at 95+10=105 meters the body stops and starts falling from rest. After 105 meters of free falling the speed will be

$$v_\text{fin}=\sqrt{2g(h+h_\text{up})}=\sqrt{2\cdot9.8\cdot105}=45.37\text{ m/s}.$$