Ignoring Friction and Aerodynamic Effects, Calculate the Speed Needed by the Skier on Leaving the Ramp?

Published July 7, 2010

Question by NA: Ignoring friction and aerodynamic effects, calculate the speed needed by the skier on leaving the ramp?
A Nordic jumper goes off a ski jump at an angle of 9.8° below the horizontal, traveling 108 m horizontally and 64.3 m vertically before landing.
(a) Ignoring friction and aerodynamic effects, calculate the speed needed by the skier on leaving the ramp.

Best answer:

Answer by DH
vx0 = v0*cos(-9.8) and vy0 = v0*(sin-9.8)

Now x = vx0*t so t = x/vx0 = 108/v0*cos(-9.8)

From the y we have y = y0 + vy0*t – 1/2*g*t^2

subbing in we get

0 = 64.3 + v0*sin(-9.8)*108/(v0*cos(-9.8)) -1/2*g*(108/v0*cos(-9.8))^2

So -58859/v0^2 – 18.655 + 64.3 = 0

So 58859/v0^2 = 64.3-18.655 = 45.645

So v0 = sqrt(58859/45.645) = 35.9m/s

Add your own answer in the comments!