91Ó°ÊÓ

This force's magnitude is proportional to the magnitude of the acceleration.

The ball has mass here.$m=50\mathrm{g}$

If the ball is simply released, there will be no air drag.

Then comes the acceleration.

$a_a=g$.

we get mass of ball is$m=100\mathrm{g}$

After that, the ball is simply released.

There will be no opposition from the air.

Then comes the acceleration$a_b=g$.

The net acceleration of the ball when it is travelling downward is

$a=-\left(g-\frac{D}{m}\right)$

The force of drag is$D=\frac{1}{4}A{v}^2$

The ball's mass is$m=50\mathrm{g}$

$=0.50\mathrm{kg}$

we have$v=20\mathrm{m}/\mathrm{s}$

$a_c=-\left(g-\left(\frac{\frac{1}{4}A(20\mathrm{m}/\mathrm{s}{)}^2}{0.05\mathrm{kg}}\right)\right)$

$\text{Magnitude of acceleration is}\left|a_c\right|=g-\left(\frac{\frac{1}{4}A(20\mathrm{m}/\mathrm{s}{)}^2}{0.05\mathrm{kg}}\right)$.

The ball's mass is now$m=100\mathrm{g}$$=0.1\mathrm{kg}$

The ball is currently on its way down.

Then$a_d=-\left(g-\frac{\frac{1}{4}A(20\mathrm{m}/\mathrm{s}{)}^2}{0.1\mathrm{kg}}\right)$

$\left|a_d\right|=g-\left[\frac{\frac{1}{4}A(20\mathrm{m}/\mathrm{s}{)}^2}{0.1\mathrm{kg}}\right]$

e) The ball's mass is now$m=50\mathrm{g}$$=0.05\mathrm{kg}$

The ball is now going upward.

The ball's acceleration is then increased$a_e=\left(g+\frac{\frac{1}{4}A{v}^2}{m}\right)$

$=-\left(g+\frac{\frac{1}{4}A(20\mathrm{m}/\mathrm{s}{)}^2}{0.05\mathrm{kg}}\right)$

The magnitude of the ball's acceleration is

$\left|a_e\right|=\left(g+\frac{\frac{1}{4}A(20\mathrm{m}/\mathrm{s}{)}^2}{0.05\mathrm{kg}}\right)$

Now, based on the information presented above,$a_e>a_a=a_b>a_d>a_c$.