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)?

Or, is the skier truly jumping onto an uphill slope?

 

[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)?

Or, is the skier truly jumping onto an uphill slope?

Here is a picture of it:

 

[rwisz]

Your approach seems to be correct, but it's possible that there was a mistake in your calculations or in the values you used for the initial velocity and angles. It's important to double check all of your numbers and make sure they are accurate. Additionally, you may want to consider using the equations for projectile motion, as the skier is essentially following a parabolic path. This may provide a more accurate and efficient solution.