Question

A body of mass m falls freely through a height h from the top of a tower. The velocity just before touching the ground is $\sqrt{\frac{3}{2}gh}$ . The air drag is:

• A. $mg$
• B. $\frac{mg}{2}$
• C. $\frac{mg}{3}$
• D. $\frac{mg}{4}$

Answer 1

The correct option is D $\frac{mg}{4}$

Let $a_{net}$ be the net acceleration

using: $v^2-u^2=2as$

$u=0,v=\sqrt{\frac{3}{2}gh},a=a_{net},s=h$

$\frac{3}{2}gh=2a_{net}h$

$a_{net}=\frac{3}{4}g$

$g-\frac{F_a}{m}=\frac{3}{4}g$

$F_a=\frac{mg}{4}$