Skateboarder Reaches 4.45m in Half-Pipe Event at Olympics

[Chocolaty]

A skateboarder is competing in the half-pipe event at the Olympics. He reaches the top of the half pipe at a vertical velocity of 10 m/s. His total mass is 70 kg.

a) Calculate the kinetic energy of the competitor when he leaves the half pipe. Ignore the effects of friction.

Ek = 0.5mv^2
Ek = 0.5*70*100
Ek = 3500J = 3.5KJ

Here's where I'm not so sure...
b) If frictional forces of 100 N were acting on the skateboarder, what height would he reach.
Okay so once he leaves the pipe the only two forces which are pulling him down are gravity and this frictional force

Fg = 70*9.8 = 686N
686N + 100N = 786N

So now i'll find the acceleration
-Ff = ma
-786 = 70*a
a = -11.23m/s^2

V2 - V1 = a*t
0 - 10 = -11.23*t
t = 0.89s

d = v1t + 0.5at^2
d = 10*0.89 + 0.5*-11.23*0.89^2
d = 4.45 meters

I have a feeling I'm doing something illegal here, or just not right.
Is this ok?

 

[Leong]

i think it is ok.
the steps make sense.

 

[Pengwuino]

Seems about right. You can check by using conservation of energy since you know the final and initial positions.