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Projectile Motion: Involves motorcycle jumping

  1. Sep 20, 2004 #1
    Hello,

    This problem has been troubling me for some time now and it's time for a little help. The problem says: A motorcycle daredevil is attempting to jump across as many buses as possible (see the drawing). The takeoff ramp makes an angle theta = 18.2 degrees above the horizontal, and the landing ramp is identical to the take off ramp. The buses are parked side by side and each bus is 2.74m wide. The cyclist leaves the ramp with a speed of v = 30.0 m/s. What is the maximum number of buses over which the cyclist can jump?

    (Note: I attached the drawing below.)

    So my thinking is that if I find the distance traveled and divide that by 2.74 then I will know how many buses will fit.

    My problem is I'm not sure exactly how to go about doing that.

    Here's what I know. I believe there is constant velocity.

    Now my variables:
    x : ?
    Ax : 0 m/s^2
    Vx : 30.0(cos 18.2 degrees)
    Vox : 30.0(cos 18.2 degrees)
    t : ?

    y : ?
    Ay : ?
    Vy : ?
    Voy : 30.0(sin 18.2 degrees)
    t : ?


    So first I think I would need to find t but without having a distance I don't have a clue how to go about this. Any help would be greatly appreciated.
     

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  3. Sep 20, 2004 #2

    arildno

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    You know that the height of the ramp equals 2.74m (or 0 if you measure height from take-off level, ratger than ground).
    By using projectile motion you may compute where other point at the same level will be.
     
  4. Sep 20, 2004 #3
    How do I know the ramp is 2.74m high? It says the width of the bus is 2.74 but says nothing about the height of the bus. And I didn't follow your last sentence.
     
  5. Sep 20, 2004 #4

    arildno

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    Oops, I'm sorry about the 2.74's :redface:
    It doesn't matter though:
    From the picture, the ramp height is on level with the buses' height's, and that's what do matter.
    You know the equations for projectile motion, right?
    Say that the level for take-off is y=0 (this is clearly admissible!).
    When you solve for the motor-bike's vertical position, it must have TWO solutions for y=0; namely, at t=0 (initially) and at some other, non-zero t-value T
    Between t=0 and t=T, the motor-bike is clearly above the buses.
    By inserting t=T in the horizontal position function, you'll find the distance D the bike has travelled horizontally.
    Dividing D with 2.74 (D/2.74) you'll find the number of buses you can squeeze in.
     
  6. Sep 20, 2004 #5
    OK, I've been reading your response over and over again, but for some reason I'm just not understanding.

    I do know the equations for projectile motion.

    I think I understand that I can use y=0 for the take off. But I don't understand why or how to even get two solutions for y=0.

    My next question then is, am I finding the time the motorcycle is in the air in the horizontal position or in the vertical position? After looking at my choices for equations to use, I have no idea which one to even use to find the time.
     
    Last edited: Sep 20, 2004
  7. Sep 20, 2004 #6

    arildno

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    Well let's start:
    The vertical component of Newton's second law reads:
    [tex]-mg=m\frac{d^{2}y}{dt^{2}}[/tex]
    where y(t) is the vertical position at time t.
    Hence, we have:
    [tex]y(t)=0+V_{0,y}t-\frac{gt^{2}}{2}[/tex]
    where [tex]V_{0,y}[/tex] is the initial velocity in the vertical direction.
    (Since you've been given initial speed, and the angle of the ramp, you may deduce from this [tex]V_{0,y}[/tex])
    From this expression for y(t), you may find the t value T not equal to zero so that
    y(T) =0
     
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