A man skis down a slope 120 m high. If 80 percent of his initial potential energy is lost to friction and air resistence, what is his speed at the bottom of the slope?

Answer 1

20% of M g H is converted to (1/2) M V^2
Note that the mass M cancels out.

0.2 gH = 0.5 V^2

H = 120 m; g = 9.81 m/s^2

Now just Solve for V

Answer 2

45

Answer 3

PE=KE
mgh=1/2mv^2
0.2(9.8m/s^2)(120m)=0.5(v^2)
(235.2m^2/s^2)/0.5=(0.5v^2)/0.5
v^2=470.4
v=21.689 m/s