# First time poster needs help with a problem at wits end

1. Feb 28, 2006

### Finns14

fter leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 51.8 m horizontally from the end of the ramp. His velocity, just before landing, is 20.7 m/s and points in a direction 43.3 degrees below the horizontal. Neglecting air resistance and any lift that he experiences while airborne, find (a) the magnitude and (b) the direction of his initial velocity when he left the end of the ramp.

That is the problem I understand mostly how to do this problem but this is only after hours of struggling with it. I am getting frustrated that the simple math and assiging negative and positive vaule are making me unable to solve the problem.

2. Feb 28, 2006

### Astronuc

Staff Emeritus
Starting with
one resolves the velocity (vector) into downward (y) and horizontal (x) components. The y velocity component involves an acceleration (increase downward) due to gravity.

The x-component does not change, it is constant during the descent because there is no acceleration in the x-direction after the skier leaves the ramp.

One has the horizontal distance traveled (x=51.8 m) at vx, and the time t = vx/x

3. Feb 28, 2006

### Hootenanny

Staff Emeritus
Welcome to PF! The first thing you need to decide is a co-ordinate system. If you use a standard x-y axis as your template, then verically up will be positive, downa will be negative and left to right movement will be positive. What you need to do first is split the velocity into vertical and horiztonal components by resolving the velocity using trig.