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Projectile Motion: Finding the correct angle for a launch

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 69.4 m across. If he desires a 3.0-second flight time, what is the correct angle for his launch ramp (deg)?


    2. Relevant equations
    vx = vxo

    x − xo = (vox)t

    (vy) = (vyo) − gt

    y − yo = (voy)t −(1/2)g t^2

    Range =((vo^2)/g) sin(2θ)

    3. The attempt at a solution
    x − xo = (vox)t
    69.4m = (vox)(3s)
    23.13 m/s = vox

    Range =((vo^2)/g) sin(2θ)
    69.4m ={ (23.13^2 m/s)/9.81 m/s^2 } sin(2θ)
    θ = error
     
  2. jcsd
  3. Sep 7, 2011 #2

    ehild

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    This formula is valid when the initial and final heights are the same. It is not true now, as there is a 15 m drop.

    ehild
     
  4. Sep 7, 2011 #3
    Didn't realize this formula had that limitation. How would you find the correct angle without it? I have no other formulas involving theta.
     
  5. Sep 7, 2011 #4

    ehild

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    You have the formulae

    x=v0xt
    y=y0+v0yt-g/2 t2.

    You know t, the horizontal distance and the height from where the motorcycle jumps (15 m) with respect to the final height (zero). Can you get v0y?
    If you have both the x and y components of the velocity, can you find its angle with the positive x axis?

    ehild
     
  6. Sep 7, 2011 #5
    I went that route at first and ended up with a wrong answer. I've checked my calculations 5 times and keep coming up with the same wrong angle so if you wouldn't mind double checking me, I'd appreciate it.

    y − yo = (voy)t −(1/2)g t^2
    15m = (voy) (3s) - (1/2) (9.81m/s^2)(3s)^2
    59.145 = (voy) (3s)
    voy = 19.7 m/s

    x − xo = (vox)t
    69.4m = (vox)(3s)
    23.13 m/s = vox

    To find the angle I used:
    tan^-1 of (19.715/23.13) = 40.44 degrees
     
  7. Sep 7, 2011 #6

    ehild

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    The problem says that the motorcycle drops 15 m. Your formula means that it raises as your final position is 15 m higher than the initial one. Correct it.

    ehild
     
  8. Sep 7, 2011 #7
    I finally see where I went wrong. I chose down as the positive direction for y to not have to deal with negative numbers, hence the positive acceleration and positive 15m.

    Where this went wrong is that the formula already factors a negative downward acceleration into it with the -(1/2)gt^2 so the downward direction has to be negative.

    Thank you so much for helping me realize this.

    I ended up with the right answer of 22.8 degrees.
     
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