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Projectile Motion - Ramp Angle

  1. Jan 12, 2008 #1
    [SOLVED] Projectile Motion - Ramp Angle

    Good evening everyone!
    1. The problem statement, all variables and given/known data
    This is a classic ramp angle question. I am trying to find the angle of a ramp in order for a bus to complete the gap (Yes, this is from the film Speed, and I have searched, but haven't found a solution.)
    Known:
    Velocity = 31.29 m/s
    Distance of gap = 15.24 m
    The landing part of the road is level with the take off.

    2. Relevant equations
    θ = (1/2) sin^-1 (fg x d / v2).

    3. The attempt at a solution
    θ = (1/2) sin-1 (9.81 x 15.24 /31.292)
    θ = (1/2) sin-1 (149.5044 / 979.0641)
    θ = (1/2) sin-1 (0.1527)
    θ = (1/2) (8.7834)
    θ = 4.3917 degrees.

    This answer does not quite seem right. I have gone through my calculations, but I cannot find an error. Perhaps my equation is wrong, I am not sure.
     
    Last edited: Jan 13, 2008
  2. jcsd
  3. Jan 13, 2008 #2
    I am still puzzled by this.
    Could someone help me determine whether or not I am using the correct equation?
     
  4. Jan 13, 2008 #3
    I see that you are using the range formula. Make sure the distance is in meters. Also remember that sin(x) = sin(180 - x), so there are two solutions for the angle that yield the same range. One of the angles is 4.39 degrees while the other angle is 85.6 degrees.
     
  5. Jan 13, 2008 #4
    Thank you. I converted the distance to meters. I doubt the angle of the ramp would have to be 85.6, so you would say 4.39 is correct?
    I was expecting a number closer to around 25 degrees, but that was pure guessing.
     
  6. Jan 13, 2008 #5
    The range equation is

    [tex]R = \frac{v^2}{g}\sin{2\theta}[/tex]

    rearranging

    [tex]\sin{2\theta} = \frac{Rg}{v^2}[/tex]

    There are two solutions for theta

    [tex]\theta = \frac{1}{2} \sin^{-1}\frac{Rg}{v^2}[/tex]
    or
    [tex]\theta = \frac{1}{2}[180 - \sin^{-1}\frac{Rg}{v^2}][/tex]

    Plugging in all the numbers, you get
    [tex]\theta = 4.39^\circ[/tex]
    or
    [tex]\theta = 85.61^\circ[/tex]

    Any value of theta between those two will make the jump (since at either of those values, the bus barely makes it)
     
  7. Jan 13, 2008 #6
    Thank you very much, sir.
    This question had me second guessing myself because we were not given the equation, I had just found it somewhere online.
    Cheers!
     
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