1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Super fun projectile motion ski slope problem!

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A ski-jumper comes off the bottom of the slope going horizontally at a speed of 20 ms. It starts off 3 m above a slope going downwards at an angle of 45 degrees. How far along the slope does it land.

    2. Relevant equations

    1) y= Vsinθ - (.5)gt^2
    2) x= Vcosθt
    3) tanβ = ||y|| / ||x||

    3. The attempt at a solution

    I attempted to find the solution by considering the beginning of the slope to be at the origin with the skiier starting 3 metres above it so my equation for y position was:

    4) y = Vsinθt - (.5)gt^2 + 3

    -which simplifies to: y = -.5gt^2 + 3 since sinθ is zero.

    The x position is given by equation 2.
    Using equation 3 to change equation 2 to give the y-position corresponding to the x-position led me to:

    -y/V = t (since Vcosθ = V in this case)

    I then used this in equation 4 to replace t, and used the quadratic formula to get an answer for the y value.

    After this step the distance is just a matter of finding the hypotenuse of a triangle and isn't of interest to me, but what I'm concerned about is that the answer I got is off slightly from what I found when I measured out the problem on computation paper and it's outside (just by a little too much) the error margins given the tools I used for drawing it.

    My question is whether or not I correctly incorporated the starting position of 3 metres above the 45 degree angle slope? Thanks :)
  2. jcsd
  3. Oct 11, 2012 #2


    User Avatar

    Staff: Mentor

    That looks right. If you'd shown your working and answers I would have seen whether your answers accord with mine.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook