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Homework Help: Projectile Motion of a cannon

  1. Oct 5, 2007 #1
    A cannon (at the origin) shoots a ball at an angle [tex] \theta [/tex] above the horizontal ground. Neglect air resistance. Let [tex] r(t) [/tex] denote the ball's distance from the cannon. What is the largest possible value of [tex] \theta [/tex] if [tex] r(t) [/tex] is to increase throughout the ball's flight? Taylor, Classical Mechanics problem 1.40

    Alright, so I have the equations for the components of the distance vector:
    [tex] x = \cos(\theta) \cdot v_{o} \cdot t [/tex]
    [tex] y = \cos(\theta) \cdot v_{o} \cdot t - \frac{g^2}{2} \cdot t^2 [/tex]

    Next I create the distance vector, take its magnitude([tex] \sqrt{x^2 + y^2} [/tex]), and square it to get:

    [tex] v_{o}^2 \cdot t^2 + \frac{g^2}{4} \cdot t^4 - g \cdot t^3 \cdot v_{o} \cdot \sin(\theta) [/tex]

    My next step would be to take the derivative of the function (with respect to time) and set the condition that time is greater than zero. This would allow me to keep the derivative monotonic. With this in mind I would find where the derivative is equal to zero and solve for time by using the quadratic formula. Then I would take the discriminant, set it to be greater than zero, and find solve for theta.

    [tex] \frac{d(r^2)}{dt} = t(2 \cdot v_{o}^2 + g^2 \cdot t^2 - 3g \cdot t \cdot v_{o} \cdot \sin(\theta)) [/tex]

    However, when I do this process the discriminant gives me the condition that [tex] \theta [/tex] is to be greater than 70.5 degrees. This is obviously wrong but no matter how many times I've looked at it today I can't find my mistake. Anybody out there willing to help me figure out what I'm doing wrong?
  2. jcsd
  3. Oct 5, 2007 #2


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    Looks right to me. Why do you think it is wrong? The question was to find the largest angle such the that distance is still increasing. Anything larger and the distance won't always increase. You found that angle.

    An easy sanity check can be done. If it is fired straight up (90 degrees), then obviously the distance won't always be increasing because it comes back to you. At a very shallow angle (~0) it is obvious that the distance will always be increasing.
    Last edited: Oct 5, 2007
  4. Oct 5, 2007 #3
    It seems wrong because my discriminant gives the relation of [tex] \theta [/tex] > 70.5 rather than less. Also, 70.5 seems too big because by that angle I would have thought that the decrease in y would have overtaken the increase in x. Lastly, my teacher didn't exactly find my answer to be correct.
  5. Oct 5, 2007 #4


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    Not convinced? If you have Excel, a not so elegant way to verify is to plot r(t) vs t (up to the value of t where y = 0) at various angles. If there is no max inbetween the values of t where y = 0, it means the distance is always increasing. You'll soon see you have the right answer. You'll have to pick an arbitrary value for [itex]v_0[/itex] but you'll quickly find that any value will provide the same result.

    By the way, you have a typo in your expression for y. It should be:

    [tex] y = \sin(\theta) \cdot v_{o} \cdot t - \frac{g^2}{2} \cdot t^2 [/tex]
    Last edited: Oct 5, 2007
  6. Sep 16, 2008 #5
    I don't understand why you squared r. Why is g(t)=r^2?
  7. Sep 16, 2008 #6
    why is the g in the equation for y squared? shouldn't g just be to the first power?
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