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Projectile motion motorcycle question

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A motorcycle daredevil wants to ride up a 50.0m ramp set at a 30.0 θ incline to the ground. It will launch him in the air and he wants to come down so he just misses the last of a number of 1.00 m diameter barrels. If the speed at the instant when he leaves the ramp is 60.0 m/s, how many barrels can be used?
    the answer is 355

    2. Relevant equations
    Vf=Vi*cosθ
    x=xo+vi*cos θ*t
    Vf=Vi*sin θ -gt
    y=y0+vi*sin θ*t-1/2gt^2
    height=Vi^2*sin θ*cos θ/2g
    Range=Vi^2*sin2 θ/g
    3. The attempt at a solution
    Vf=Vi*cos30
    Vi=60 / cos30
    Vi=69.3 m/s

    R=Vi^2*sin2 θ/g
    R=69.3^2*sin60/9.8
    R=424

    I
     
  2. jcsd
  3. Feb 20, 2016 #2

    haruspex

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    That's a new one on me. What do you mean by vi and vf there?
     
  4. Feb 20, 2016 #3
    I find it on my book
    It should be
    Vx=Vi+at for x component
    (since a=0)

    Vx=Vi*cosθ
     
  5. Feb 20, 2016 #4

    cnh1995

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    You need to sketch a diagram first. First calculate the time after which the rider will land. Use that time to find the horizontal distance travelled i.e. d=Vprojectioncosθ. No of barrels will be simply the horizontal distance since the diameter of each barrel is 1m.
     
  6. Feb 20, 2016 #5

    haruspex

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    That makes more sense. Which of those velocities is given?
     
  7. Feb 20, 2016 #6
    vx is equal to 60m/s?
    I try to sketch this diagram
    5817VS.jpg
     
  8. Feb 20, 2016 #7

    cnh1995

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    Vx is not 60m/s. It is the total velocity of projection, inclined at 30 degrees with the horizontal. That's what you are supposed to show in your diagram.
    Now, from your diagram, you can see the net vertical displacement of the rider. You know the acceleration due to gravity g. Set up an equation relating displacement with time and g and find the time taken by the rider to reach the ground. You can then use this time to find the horizontal distance covered before landing.
     
    Last edited: Feb 20, 2016
  9. Feb 20, 2016 #8
    But how to use projectile velocity find the Vi?
    50*sin30=Vi*sin30*t -1/2gt^2
    x=x0+Vi*cos30t
    these equation?
     
  10. Feb 20, 2016 #9

    cnh1995

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    Right! You have Vi already.
     
  11. Feb 20, 2016 #10
    Vi equal to 60?
     
  12. Feb 20, 2016 #11

    cnh1995

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    Yes.
     
  13. Feb 20, 2016 #12
    But I substitute vi=60 to
    50*sin30=Vi*sin30*t -1/2gt^2
    I get t=0.995s or 5.127s
    then put the t =5.127 into
    x=x0+vi*cos30*t
    then x=266
    what's wrong with it?
     
  14. Feb 20, 2016 #13

    cnh1995

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    Well, the displacement should be negative i. e. -25m.
     
  15. Feb 20, 2016 #14

    haruspex

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    No. He leaves the ramp at a speed of 60m/s. At that time, he is already at a height of 50 sin(30) m.
     
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