Question

A $50\text{kg}$ skydiver falls through the air and attains terminal velocity after some time. The drag force is a function of velocity given as $F_{\text{drag}}=-b{v}^2$ where the negative sign denotes that the drag force is opposite to the direction of velocity. What is the terminal velocity of the skydiver (assuming the drag constant $b$ is $0.2\text{kg/m})$?

• A. $5\text{m/s}$
• B. $250\text{m/s}$
• C. $50\text{m/s}$
• D. $100\text{m/s}$

Answer 1

The correct option is C $50\text{m/s}$

When the body falling through the air attains terminal velocity, the drag force$\left(F_D\right)$ will be equal to the weight of the body.
$\because a=0$, when body reaches terminal velocity.
And density of air is too low for considering buoyant force.
$\Rightarrow F_D=mg$
or, $b{v}_T^2=mg...\left(i\right)$
[Magnitude of drag force is $b{v}^2$]
Terminal velocity $v_T=\sqrt{\frac{mg}{b}}=\sqrt{\frac{50\times 10}{0.2}}$
$\therefore v_T=50\text{m/s}$