An airplane is flying with a speed of 250km/h at a height of 3610m above the ground. A parachutist whose mass is 75.2kg, jumps out of the airplane, opens the parachute and then lands on the ground with a speed of 3.75m/s . How much energy was dissipated on the parachute by the air friction?

Answer 1

So I assumed the horizontal speed of 250km/h of the airplane is not needed to obtain the solution?
Using Conservation of Energy theorem:
Epgi+Eki = Epgf+Ekf+E
mghi+(1/2)mvi^2 = mghf+(1/2)mvf^2+E
(75.2kg)(9.80m/s^2)(3610m)+(1/2)(75.2kg)(0m/s)^2 = (75.2kg)(9.80m/s^2)(0m)+(1/2)(75.2kg)(3.75m/s)^2+E
2660000J+0J = 0J+529J+E
E = 2660000J-529J
= 2660000J

(The full answer I got without rounding was 2659896.85J), but apparently this is incorrect? How?

Answer 2

His initial KE was dependent on air speed,

Initial energy=energy dissipated+final KE
1/2 mvi^2+mgh=energy dissipated +final KE.

In yours, you assumed vi=0, not true. It is the initial airspeed. You have to assume the parachute absorved it.

Answer 3

Thank you. But I am still confused about why exactly the airplane's speed affects the parachutist's initial kinetic energy. The way I break it down is that the airplane's speed is horizontal and the parachutist's initial kinetic energy deals with vertical components.

Answer 4

KE is NOT A Vector, nor is it directional.