Kinematics: Final velocity components

[huynhtn2]

Homework Statement

A skier leaves the ramp of a ski jump with a velocity of 10.9 m/s, 13.9° above the horizontal. The slope is inclined at 48.4°, and air resistance is negligible. Calculate the velocity components just before the landing. Enter the x-component (horizontal) first and then the y-component (vertical).

Homework Equations

d=vit + 0.5at^2
vf=vi +at
vf^2=Vi^2 +ad

The Attempt at a Solution

This is a second part of the question. So previously i found the total time the skier was in the air which was 2.965 seconds. I took his initial velocity and did Vsintheta to find verical velocity, and Vcostheta to find the horizontal velocity. Then i found the time where the skier reached his max height by using vf=vi +at and multiplied this time by 2.

Then i subtracted this time by the total time to see the time that he will accelerate down. I added the vertical velocity with the acceleration times the time accelerating to find the final velocity. But in the end it was wrong. What did i do wrong?

 

[zgozvrm]

The slope is inclined at 48.4°

Do you mean that the slope is declined at 48.4[itex]^\circ[/tex] (or that the slope falls at an angle of 48.4[itex]^\circ[/tex] below horizontal)?<br /> <br /> Or, is the skier truly jumping onto an uphill slope?[/itex][/itex]

 

[huynhtn2]

Heres a picture of it:

 

[huynhtn2]

Do you mean that the slope is declined at 48.4[itex]^\circ[/tex] (or that the slope falls at an angle of 48.4[itex]^\circ[/tex] below horizontal)?<br /> <br /> Or, is the skier truly jumping onto an uphill slope?[/itex][/itex]

[itex]$<br /> Here is a picture of it:$[/itex]