Question

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

• A. $5\frac{m}{s}$
• B. $50\frac{m}{s}$
• C. $100\frac{m}{s}$
• D. $250\frac{m}{s}$
• E. $2500\frac{m}{s}$

Answer 1

Answer: (B)

$50\frac{m}{s}$

When the falling body through the air will reach to the terminal velocity, the drag force will be equal to weight of the body.

Thus, $b{v}^2=mg$

The terminal velocity, $v=v_t=\sqrt{\frac{mg}{b}}=\sqrt{\frac{50\times 10}{0.2}}=50m/s$

Thus , option B will be correct.