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Homework Help: Projectile motion skateboarder problem

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A skateboarder starts up a 1.0-m-high, 30-degree ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction. How far from the end of the ramp does the skateboarder touch down?

    angle = 30 degrees
    vi = 7.0 m/s
    a = -9.8 m/s2
    initial height = 1 m

    2. Relevant equations
    y = x*tanθ + ax2/(2*vi^2*cos2θ) [graphs the path of a projectile]


    3. The attempt at a solution
    First, I graphed the path of the projectile, and since it forms a parabola whose roots are the max distances, I got an answer of 5.65 m. The homework problem is online, so it told me that it was wrong.

    Not knowing where I went wrong graphing it, I found the time it spent in the air by plugging into vf2=vi2+2ad, and then plugging vf into vf=vi+at for t0 to the max height. Then I just used the diy= .5at2 equation to find the time form the peak to the ground. The total time I got was .933 seconds. Multiplying this by the horizontal component of his velocity (vi*cosθ = 7.0*cos30 = 6.1) I got 5.69 seconds (probably a little different from my calculator since my calculator ignores significant figures even more than I do). Again, this was wrong.

    So I gave up and clicked show answer: 3.78 meters.
    I don't get how my teacher got this answer.
    Last edited: Sep 23, 2009
  2. jcsd
  3. Sep 23, 2009 #2


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    What did you use for the initial speed for the projectile part of the motion? I hope it was not 7 m/s because that's his velocity at the bottom of the ramp. At the top of the ramp, when he is about to take off, his velocity will be less than 7 m/s and you need to calculate what it is.
  4. Sep 23, 2009 #3
    Ah, all the other problems in the set listed vi as the velocity the object becomes airborne with, so I assumed that was the case here as well. Factoring in that negative acceleration up the ramp got me 5.4 m/s leaving the ramp, which got me 3.78 meters.
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