Calculating Power of a Skydiver at Terminal Speed

[koky0]

60-kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?

can we say p=w/t=mgt/t=60*10*50/1=30000 watt!

 

[Q_Goest]

Hi Koky, welcome to the board.
Power is force times distance per unit time, so where you say P=w/t=mgt/t, that should be P=wL/t=mgL/t
Where P=power
w=weight
L=length or distance
t=time
But yes, you got it correct in the end. This begs the question, "Where is all that energy going?" Is it going into heating the air or the skydiver or both? And even if (of course it is) going into both, wouldn't the skydiver feel thousands of watts of power being blown at him? After all, a 1000 watt hair dryer is quite warm, so what should the skydiver feel and why?

 

[koky0]

Thank you Q_Goest .. you r right .. it is my typing mistake .. it should be d (distance) not t