91Ó°ÊÓ

The ratio of air density outside the balloon to that inside is.$\frac{{\mathrm{ÒÏ}}_{out}}{{\mathrm{ÒÏ}}_{in}}=1.39$

Newton’s second law states that the net force$\stackrel{→}{F}$ on a body with mass mis related to the body’s acceleration $\stackrel{→}{a}$by,

$\stackrel{→}{F}=m\stackrel{→}{a}$

We can find the acceleration of the rising hot-air balloon by applying Newton’s second law to it. Mass can be written in terms of volume and density. Using the ratio of the density of the air inside and outside, we can find the acceleration of the balloon.

Formulae:

$$\begin{gathered} F_{net}=ma \\ m=\mathrm{ÒÏ}V \end{gathered}$$

Free body diagram:

Applying Newton’s law to the rising hot-air balloon gives,

$\begin{array}{rcl}{F}_b-mg& =& ma\\ {\mathrm{ÒÏ}}_{out}Vg-{\mathrm{ÒÏ}}_{in}Vg& =& {\mathrm{ÒÏ}}_{in}Va\\ a& =& \frac{\left({\mathrm{ÒÏ}}_{out}-{\mathrm{ÒÏ}}_{in}\right)g}{{\mathrm{ÒÏ}}_{in}}\\ & =& \left(\frac{{\mathrm{ÒÏ}}_{out}}{{\mathrm{ÒÏ}}_{in}}-1\right)g\\ & & \end{array}$

Substitute the values in the above equation.

$\begin{array}{rcl}& =& \left(1.39-1\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ & =& 3.82{\text{m/s}}^{\text{2}}\\ & & \end{array}$

Therefore, the acceleration of the rising hot-air balloon is $3.82{\text{m/s}}^{\text{2}}$.