# A parachutist with a mass of 80 kg is falling at 55km/h when her parachute opens. She then falls 24 m in the next 2 s. What is the force of air resistance acting on the parachute during those 2 s?

Jul 12, 2018

$\text{912 N}$

#### Explanation:

$\text{55 km/h" = 55xx5/18 \ "m/s" = "15.2 m/s}$

This is the velocity at the time of parachute opens.

Take this moment as $t = \text{0 s}$. Hence at $t = \text{2 s}$, the displacement of the parachutist is $S = \text{24 m}$.

There will be acceleration due to gravity '$g$' and the retardation by the air resistance. Let it be $' a '$.

So the effective acceleration will be $g - a$ because the direction of $g$ is along the motion of the parachutist and the direction of $a$ is opposite to the direction of motion of parachutist.

$S = u t + \left(g - a\right) {t}^{2}$

Here initial velocity is $u = \text{15.2 m/s}$. Substituting the values we get

"24 m"=("15.2 m/s" xx "2 s")+("9.8 m/s"^2 - a)xx("2 s")^2

Solving this we get the value of $' a '$ as ${\text{11.4 m/s}}^{2}$.

Force $\left(F\right)$ is mass $\times$ acceleration, so

$F = \text{80 kg" xx "11.4 m/s"^2 = "912 N}$

The force of air resistance will be $\text{912 N}$ .