Posted by Casey on Saturday, June 18, 2011 at 4:39pm.
A parachutist with mass 79.7 kg jumps from an airplane traveling at a speed vi= 112 km/hr at a height H = 2570 m. He lands with a speed of vf = 5.21 m/s. What is the change in mechanical energy (J) of the earth parachutist system from just after the jump to just before landing?
My work: I started by solving Ug=mgh
(19.7kg)(9.8)(2570) = 20073224.2
Initial Kinetic + 1/2mgv^2
Ug + ki = Ei
final e = 1/2mgv^2f
change in E = EiEf
The problem I am having is with the simple calculations. When I plug in my mass in kg and velocity in m/s.. I am not getting the correct answer in Joules.. Is there some type of conversion that I am missing?

Physics  bobpursley, Saturday, June 18, 2011 at 5:10pm
I do not understand your work.
HE started with Kinetic energy and potentail energy. He ended with some kinetic energy.
change= intial energyfinal energy
= 1/2 m Vi^2 + mg*h 1/2 m vf^2
where Vi=112000/3600 m/s
Vf= 5.21m/s
m=79.7kg
I have no idea what conversions you might have done.