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Homework Help: Extra Credit- Taco Cannon Projectile (how do I find the function of the curve?)

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data
    In Austin, Texas there is a Taco Cannon (modified T-shirt cannon) that will be spreading the joy of taco'ey goodness to festival goers. The only information I have is that it can fire 200 feet (60.96 meters).

    My physics professor allows us extra credit for applying physics knowledge to the world. I want to solve for the possible area in the sky that the tacos can fill if a cannon is fired 360 degrees. I'll apply a Standard Deviation (SD) of (+/-) 5 meters to the 60.96 meters

    I'm calculating the velocities needed to achieve that range, 60.96 meters, at 45 degrees and 55 degrees.

    for 45 degrees -5m SD
    V(initial)= 23.418 m/s
    Max Height= 13.99 m
    Δ distance= 55.96 m
    time= 3.38 sec

    for 45 degrees +5m SD
    V(initial)= 25.424 m/s
    Max Height= 16.49 m
    Δ distance= 65.96 m
    time= 3.67 sec

    With the SD, I have
    2. Relevant equations

    3. The attempt at a solution

    1st I calculated the area on the ground a taco desiring person would be.
    Area of big circle minus area of small circle= (1219.2)(pi) meters squared)

    I want to calculate the 3 dimensional air space that a taco may be in at a given time (revolutions of a solid), so I can warn migrating birds of raining tacos. I'm not sure of how to get the function of the trajectory as it arcs through the air. Other than choosing points and making the curve from those data points, is there an easier way? I'm taking an algebra based physics course and we only are working with x and y coordinate independently.
  2. jcsd
  3. Nov 1, 2012 #2
    I am unsure what area you want to find. Is that the area of the "dome" formed by all the possible trajectories?

    Secondly, I think you should interpret the 200 feet given as the maximum range, which is achieved at a particular angle of elevation, from which you should be able to compute the muzzle velocity.
  4. Nov 1, 2012 #3
    I apologize for my lack of clarity. Yes, the dome is what I'm trying to calculate. I will heed your advice and go with the 200 foot maximum range (60.96m) and keep the -5m SD. The muzzle velocity is listed as the V(initial) numbers.

    for 45 degrees -5m SD
    V(initial)= 23.418 m/s
    Max Height= 13.99 m
    Δ distance= 55.96 m
    time= 3.38 sec

    45 degrees to maximum distance:
    V(initial)= 24.441 m/s
    Max Height= 15.24 m
    Δ distance= 60.96 m
    time= 3.53 sec

    Is there a way to find the upper and lower trajectory paths as functions? Is this solving for a parabolic equation?
  5. Nov 1, 2012 #4
    I am still confused by the relevance of deviation.

    The dome is obviously a figure of revolution. It is obtained by rotating the envelope of all the trajectories whose launch angle varies between 0 and 90 degrees.
  6. Nov 1, 2012 #5


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    Staff: Mentor

    Assuming a constant launch speed v, for a given value of launch angle θ you can write the equations of motion y(t) and x(t). That means you can eliminate t and find the trajectory equation y versus x, or y(x) for that value of θ. That is, there is a family of trajectory equations y(x,θ).

    For any given x value, then, you want to find the maximum value of y(x,θ) for all possible values of θ. This will give you a point (x,y) on the envelope that encloses the family of curves.
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