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?

1 Answer

Answer:

#"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"# .