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Quadratic Projectile

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Motorbike crossing a river using a ramp. River is 40m wide, the bank on the opposite side is 15m lower. The ramp angle is 53 degress. What is the speed needed at take off?

    2. Relevant equations
    I am using s = ut for the horizontal
    and s = ut x 1/2 at2
    I think because the bank is lower the other side I need to use the quadratic equation?


    3. The attempt at a solution
    Using s = ut I get ut = 50/cos 53
    Which I then substitue in s = ut x 1/2 at2
    Which gives:
    15 = 40/cos53 + 1/2 at2
    Which i rearrange:
    4.92 + 40/cos53 -15 = 0
    I then try and use the quadratic equation:
    t = (-66√662-4 * 4.9 * -15) /9.8
    Which gives 0.22s or -13s

    Am I anywhere near the solution?



    I know the answer is 19.8 m/s
     
    Last edited: Feb 4, 2012
  2. jcsd
  3. Feb 4, 2012 #2
    Are you sure the answer is 19.8 m/s?
     
  4. Feb 4, 2012 #3
    Yes, the answer is in my textbook
     
  5. Feb 4, 2012 #4
    Check your directions.Gravity acts downwards, you've substituted g without a negative sign.
     
  6. Feb 4, 2012 #5
    t = (-66√662-4 * 4.9 * -15) /9.8
    So I use -4.9 instead
    t = -66 + or - ( √662 -4 * -4.9 * -15) /9.8
    t = -66 + or - ( √4356 - 294) /9.8
    t = -66 + or - (66-294) /9.8
    t = -66 + or - -23
    t = -66+-23 = -89 or -43

    t must be +ve?
     
  7. Feb 4, 2012 #6
    The bank on the other side is 15 m LOWER. So the displacement is actually "-15" m.

    You don't have to choose the conventional directions like g acts downwards, but whatever you choose, make sure you do choose something, and follow it throughout the question. If "up" is positive, it should be positive throughout.
     
  8. Feb 5, 2012 #7
    Just checked the answer and its actually 17.8 ms-1, will try and work through it again later.
     
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