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Firing cannon

  1. Oct 29, 2004 #1
    A cannon fires a ball at 1000m/s. It needs to hit a target 2000m away horizontally and 800m away vertically. What angle should the cannon be fired at?

    I used the equation h=v^2*sinx^2/[2g]

    800=1000^2 * sinX^2 / [2(9.8)]

    however, I gt 7.2 after I crunch out the numbers.
    The answer in to book says about 22 degrees.

    what am I doing wrong?
  2. jcsd
  3. Oct 29, 2004 #2
    I think you better check your text book.

    The equation you used is only for finding the horizontal range not for the vertical height.

    To solve the problem, you should really be looking at the formula of projectile motion.
  4. Oct 29, 2004 #3
    All you've been told is that it must hit a target at a distance 2000m away and 800 m high. What makes you think that these paramaters necessarily correspond to the maximum height (and therefore the 2000m to half the range)?

    Suppose the firing angle is [tex]\alpha[/tex] then the equation of trajectory of a projectile is

    [tex]y = x\tan\alpha - \frac{1}{2}\frac{gx^2\sec^2\alpha}{u^2}[/tex]

    Now do you know where you went wrong in your reasoning?

    Hope that helps...
  5. Oct 30, 2004 #4
    y=800m ?
    x=2000m ?
    v=1000m/s ?

    it doesn't come out equal
  6. Oct 30, 2004 #5
    u know the distance in the x direction.

    u know the distance in the y direction.

    You know the initial velocity.

    Find the horizontal and the vertical velocities.

    Try to figure out a formula for finding the angle.

    This should be in your book.

    Use this formula to find the angle.

    Hope this helps
  7. Oct 30, 2004 #6
    Velocity in the x -direction. v0 * distance in the x direction

    Velocity in the y direction v0 * distance in the y direction

    tan theta = Vy/Vx

    theta is inverse tan of Vy/Vx
  8. Oct 30, 2004 #7
    yeah, but how do I take account for gravity?
  9. Oct 30, 2004 #8
    I'm not sure if this is correct, but I was thinking maybe you should use a system of equations. Since you know the x-component of the velocity vector is 1000*cos(theta) and the y-component is 1000*sin(theta), you can use a system of equations to solve, I think, by substituting those above values in for [itex]v_o[/itex] in their respective equations. There's likely an easier method, though..

    Here's a start:

    [tex]1000cos(x)t = 2000[/tex]
    [tex]1000sin(x) - 4.9t^2 = 800[/tex]

    Hope that helps!
  10. Oct 30, 2004 #9
    WOW, that took amazing long. But I got the answer...actually 3 worked. This is insane. Is there a faster way???

    By they way, the equation should not be 1000sin(x)-4.9t^2=800, it should have been 1000sin(x)t-4.9t^2=800

    (d=Vit+.5at^2)<-- you forgot the t after 1000sin(x) :rolleyes:
    Last edited: Oct 30, 2004
  11. Oct 30, 2004 #10
    I'm curious what you did to get the final solution. I got the equation
    2000tan(X)-19.6/cos^2(x)-800 and had to graph it to get the solution. It worked out but I never like having to graph things to get it.
  12. Oct 30, 2004 #11
  13. Oct 30, 2004 #12
    Heh, I'm really sorry you had to solve that...crazy mess of a system it is. And I forgot about the "t," doh! :) I'm sure there's a better way, but I can't think of one.
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