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Homework Help: Motorcycle jump with a twist

  1. Feb 3, 2014 #1
    Problem: A motorcyclist is about to jump across a river. The takeoff ramp is inclined at 53 degrees above the horizontal, the river is 40.0m wide, and the far bank is 15.0m lower than the top of the ramp. Ignore air resistance. What must the motorcyclist's speed be at the top of the ramp in order to just make it to the edge of the far bank?

    So I know that in the x-direction: a=0, v=initial velocity * cos(53), position=Vi*cos(53)*t
    and in the y direction: a=-9.8, V=-9.8*t+Vi*sin(53), position = -4.8t^2+Vi*sin(53)+15 (if "origin" here is (0, 15).

    So I'm missing time and initial velocity, but the equation for velocity in the x direction doesn't have t in it, so I know V=40/t, and so Vi=40/(t*cos(53)). Then I can plug that equation in for Vi in the y-position equation and solve for t (so I have t= āˆš(((40*sin(53)/cos(53))+15)/4.8) so t=3.77 seconds, and plug that value into either velocity equation, and I get 10.61m/s. Only problem is, this is the wrong answer, and the correct one is 17.8m/s - so where am I going wrong here? I've looked at a couple of explanations of this problem not involving an overall change in y and I understand why this should work, but I've redone and gone over the problem multiple times now and there is no 17.8m/s to be found!

    so I have t= āˆš(((40*sin(53)/cos(53))+15)/4.8
    Last edited: Feb 3, 2014
  2. jcsd
  3. Feb 3, 2014 #2
    Correction - because I get (40*sin(53)*t)/cos(53)*t, the t's cancel so I didn't use my calculator, I used algebra. Which is generally where my errors enter!
  4. Feb 3, 2014 #3


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    4.9 m/sĀ² ... and I think your velocity term didn't get its sign reversed.
  5. Feb 3, 2014 #4
    Oops on the 4.9 - but even changing that my answer is still far from correct. Which velocity term do you mean? -4.9t^2 just goes on the other side and become positive so all terms are positive.
  6. Feb 3, 2014 #5


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    I get t = 3.73, but even if I take t=3.77 and plug it into your Vi*cos(53)*t equation I get a number quite close to the answer. Please post all your working.
  7. Feb 4, 2014 #6
    Aha! It's my algebra. I was using the original Vx = 40/t, but NOT using the entire expression, which is (40/t)/cos(53) = 17.63 I guess 40/t gives you the average velocity, but that is not equal to the initial velocity - right? I also tried plugging in t=3.77 back into the Vy equation, but I got a huge number - I'm not sure exactly why that wouldn't work, except that t=3.77 is the "zero" of that equation.

    Thanks for your help - hopefully now that I know where I went wrong this type of question will seem as easy as it should be (since we're moving rapidly past this type of problem in class now.)
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