Skydiver accelerates towards the ground at at 9.81ms ^-2

[jimmer]

Hi,


A skydiver accelerates towards the ground at at 9.81ms ^-2. At the instant that je leaves the aeroplane.
A) The skydiver opens his parachute. Explain why he reaches a terminal velocity shortly afterwards.

If you could help with that question be very greatful
Thanks in advance
JImmer

 

[Molydood]

terminal velocity occurs when your upward forces (due to air friction and dependant on your speed and surface area) equal your downward accelerative forces (due to gravity)
as your speed or surface area increase, so do your upward (decelerative) forces
opening a 'chute increases air resistance and thus upward force, and thus equalising the upward and downward forces sooner than if you were falling without a 'chute

in a nutshell, terminal velocity for a parachute+man is much lower than terminal velocity for a man without parachute, hence you reach that speed sooner (you could even already be at the 'chute+man terminal velocity when your 'chute opens)

 

[Bob S]

High velocity air drag is turbulent drag (high Reynolds number) so the upward drag force is proportional to the square of velocity, and determines the terminal velocity (when upward and downward forces are equal). See
http://en.wikipedia.org/wiki/Drag_(physics )
When a skydiver spreads his/her arms out, the drag increases and the terminal velocity decreases.
Bob S