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?