Question

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?

Answer 1

`"912 N"`

Explanation:

`"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="0 s"`. Hence at `t = "2 s"`, the displacement of the parachutist is `S = "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=ut+(g-a)t^2`

Here initial velocity is `u = "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 `"11.4 m/s"^2`.

Force `(F)` is mass `xx` acceleration, so

`F="80 kg" xx "11.4 m/s"^2 = "912 N"`

The force of air resistance will be `"912 N"` .