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Projectile Motion of a Skier Problem

  1. Oct 23, 2010 #1
    This is the problem given:
    A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s east and lands 70.0 meters from the base of the ramp. How high is the end of the ramp from the ground?


    Relevant Equations:
    Vy=Vsin
    Vx=Vcos
    T=(2Vsin/g)
    H=(Vsin)^2/2g
    R=2V/g(sin)(cos)
    X=Vi(T) + (1/2)a(t)^2



    My attempt

    I drew a picture with the ramp horizontally, a vertical line coming off of the end of it for the height, and a parabolic curve by the vertical one to represent the path the skier took.

    then I wrote what I know:

    a= -9.81 m/s
    d=70.0 m
    t= ?
    V initial =25.0 m/s
    V final = ?

    And now I really have no idea how to find the height of the jump.
    Finding the time wouldn't help because time shouldn't affect the height of the jump. And finding the velocity wouldn't help either because velocity shouldn't affect the height of the jump either. And it doesn't seem like I can use sin/cos/tan equations because I don't have a right triangle.

    so any help is appreciated thanks! I'm at the end of my rope!
     
  2. jcsd
  3. Oct 23, 2010 #2
    It's good that your relevant equations are seperated into x and y components. But in your attempts, I don't see the same discipline. Maybe that's why you're stuck/confused.

    Consider these questions:

    x plane:
    Given velocity is 25, distance 70. Is there any accleration? If yes/no, can you find the total time taken? Is this time taken the same as the time taken for the vertical motion as well?

    y plane:
    What's the initial vy? Is there an equation that can solve for displacement without using final vy?

    I got 38.5m, if you need to check solution.
     
  4. Oct 23, 2010 #3
    thanks for trying to help! I have no idea what you are saying, but thanks for trying! :redface:
     
  5. Oct 23, 2010 #4
    I would really love to give you the full solution but I'm unable to due to forum rules. Perhaps I can try to be more clear:

    For all projectile motions, they're seperated into x and y components for calculations. x and y being horizontal and vertical respectively. We use equations of kinematics to calculate each component seperately.

    So for example, if you want to calculate the verticle component, you know that:
    acceleration = g = 9.81
    initial velocity for free fall = 0

    Thus, if you have time taken, you can use the equation:
    s = ut + 0.5at2, s being displacement, u being initial velocity (0) and a being your freefall acceleration.

    When you calculate for any verticle variable, you completely disregard all horizontal components. So you don't care about the horizontal distance, speed or acceleration. Just the verticle values alone.

    When you're done, do the same for the horizontal component.

    Hope that helps.
     
  6. Oct 24, 2010 #5
    Hey! that actually kinda makes sense! Thanks!!!
     
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