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Projectile Motion and Human Cannonball

  1. May 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi,
    I am doing an end of the year calculus presentation on projectile motion in human cannonballs. As an introduction to the projectile motion equations, my group found this problem:
    The original problem with a figure of it can be found http://higheredbcs.wiley.com/legacy/college/anton/0471482730/calc_horizons/blammo.pdf" [Broken]

    I'm not a physics student, so thank you for any help!​


    2. Relevant equations
    I'm sorry, but I couldn't figure out the latex equations...Hope these are clear enough.

    Projectile Motion Formulas
    R=(vo2/g)sin(2q)
    r(t)=v0xti+(voyt-.5gt2j
    which, if I'm correct (??) breaks down into:
    x(t)=xo+voxt
    y(t)=yo+voyt-.5gt2

    (Are there equations I'm missing for this problem that would be more useful/better than these??)

    3. The attempt at a solution
    Okay...Here we go:

    R=(vo2/g)sin(2q)
    For the range of the cannon (R) to be maximum, the angle of elevation (q) must equal 45o.
    R=(352/9.798)sin(2*45)
    R=125.026m​

    x(t)=xo+voxt
    y(t)=yo+voyt-.5gt2
    @45o, vox=voy.
    vox=vocos(q)
    vox=voy=24.749m/s

    I placed the wall of fire at x=0. So, solving for xo would give the distance from the wall the cannon must be placed. (Correct??)
    0=xo+voxt
    t=-xo/vox
    Sub t into the y(t) equation:
    y(t)=20=5+voyt-.5gt2
    15=-xo-4.899(-xo/24.749)2
    xo=-17.43m

    R+xo=length of pool
    The pool is 107.596m long.​

    So, I think I have solved this correctly. One thing I'm worried about is the assumption of q=45o. I know this makes R the greatest, but does it make the length of the pool the greatest? If the angle was steeper, could the cannon be placed closer to the fire, making the pool be a greater part of the path?

    Any help or confirmation of my answers would be greatly appreciated. Thank you so much!!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 4, 2009 #2

    nvn

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    micnike1: Your equations and math are correct (for q = 45 deg), except g appears to be inaccurate. However, your equations are not general, because they assume voy = vox, which is not true in the general case. Furthermore, your worry is well founded, because your answer is incorrect. After making your equations general (as mentioned above), how about if you check your assumption for q? Try plugging in a larger value for q, but less than 53 deg, and see what happens.
     
  4. Jun 5, 2009 #3
    Thanks nvn.

    I've spent about an hour now trying to manipulate the equations into a general form. But I always end up with two variables (q and xo) and am not able to substitute one for the other. Am I missing something here? Maybe I've misunderstood what you mean...
     
  5. Jun 5, 2009 #4

    nvn

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    Your variables sound fine, if you did not use voy = vox in your new derivation. Did you? In post 1, it appears you said voy = vox.
     
  6. Jun 5, 2009 #5
    You're right, I did use voy=vox to cancel things out in the step 15=voy(-xo/vox)-.5g(-xo/vox)^2. Not using that assumption, I get down to 15=-xotan(q)-.5g(-xo/(vocos(q)))2.

    Where do I go from here. I'm like completely lost in this problem now...
    Thanks again,
    Mike
     
  7. Jun 5, 2009 #6

    nvn

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    Excellent, micnike1. Also rederive your first equation under relevant equations in post 1. Then proceed as you did in post 1 to solve the problem. Also see post 2.
     
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