# A bag of ballast is accidentally dropped from a balloon which is stationary at an altitude of 4900m. How long does it take for the bag to hit the ground?

Mar 5, 2018

$31.6$ s

#### Explanation:

A good equation to know for calculating time of falls due to gravity alone is:

$h = \frac{1}{2} g {t}^{2} \to t = \sqrt{\frac{2 h}{g}}$

Here, we can use $h = 4900$ m. There is no gravity mentioned, so I will simply use $9.8$ m/${s}^{2}$.

$t = \sqrt{\frac{\left(2\right) \left(4900\right)}{9.8}} = 31.6$ s

Mar 5, 2018

Use the equations of motion to answer the question.

#### Explanation:

It is given that balloon is stationary at height $\left(s\right) = \text{4900 m}$. The initial velocity $\left(u\right) = \text{0 m/s}$.

As it falls in the downward direction, it accelerates with an acceleration of $g = {\text{9.81m/s}}^{2}$ (for simple calculations, $g = {\text{9.8 m/s}}^{2}$)

So

$4900 = 0 \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

On solving

$t = \sqrt{\frac{2 \cdot 4900}{9.8}}$

$t = \sqrt{1000} \text{ s}$