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General case for pojectile motion from height h

  1. Nov 2, 2005 #1
    Problem: A cannon that is capable of firing a shell at speed V is mounted on a vertical tower of height h that overlooks a level plain below.
    (a) show that the elevation angle a at which the cannon must be set to achieve maximum range is given by the expression:
    csc^2 a = 2*(1 + g*h/v^2)
    I came up with the parabolic equations
    x(t) = v*t*cos a
    z(t)= v*t*sin a - (1/2)*g*t^2 + h
    I found the time that the projectile is in the air by setting z=0 and solving for t. Then I plugged that t into the equation for x to get the range so that
    range= r = v*cos(a)*(v*sin a + sqrt( sin^2 a + 2*g*h/v^2 )
    I then differentiated r with respect to the angle a and set the result equal to zero.
    But here I am stuck. My procedure seems right to me but the algebra is gross and I can't seem to simplify it down to the form that they are asking for.
    Thanks in advance for your help.
    Last edited: Nov 2, 2005
  2. jcsd
  3. Nov 2, 2005 #2

    Maximum time in the air is obtained with elevation 90 deg (on horizontal ground). This does not give maximum range.
  4. Nov 2, 2005 #3
    Thanks so much for a quick reply.

    I thought I was supposed to find z in terms of x and then maximize x with respect to a and then solve to the form that the question is asking for. Is this incorrect? I still don't see it.
    Last edited: Nov 2, 2005
  5. Nov 2, 2005 #4


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    Homework Helper

    Your procedure is correct and the algebra is gross (I've already been there !)
    There is a lot of tedious algebra and a lengthy expression/equation. You should start to see some light on it when you manage to square the square root sign. After that you will find bits cancelling out until you get the nice final answer.

    Hint: you might find the expression easier to manipulate and visualise if you use a small trig substitution, e.g. u = sin²α. I know it helped me.
  6. Nov 2, 2005 #5
    ok so i did the substitution and substituted t into x and i've simplified is to

    range = (v^2 *sin(2a)/g)*(1+ (1+2*g*h/(v^2 *sin(a))^1/2

    is this the point at which i should take the derivative (w/ respect to a) and set equal to zero or is there a further simplification that i can make that i don't see?
  7. Nov 3, 2005 #6


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    Homework Helper

    It's more or less at this point, yes, where you take the derivative. But there are a few typos in your expression.

    1) the sin(2a) term should be sin(2a)/2 {sin(a)cos(a) = (1/2)sin(2a)}
    2) there should be a closing bracket after the power "^1/2" at the end of your expression.
    3) the V^2*sin(a) term should be V^2*sin^2(a).

    My expression was similar. It was,

    [tex]R = \frac{V cos\alpha}{g}\left(V sin\alpha + \sqrt{V^2 sin^2\alpha + 2gh}\right)[/tex]

    But you can use either. It will still be lot of work. I don't think it's worthwhile simplifying any more. Watch out for silly mistakes!

    I had a cos instead of a sin right at the beginning, and it was only when I got to about the 2nd-last line that I saw that it wasn't going to work!!
    Last edited: Nov 3, 2005
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