Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile motion, need to find angle

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A cannon with a muzzle speed of 1009 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 795 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

    2. Relevant equations
    I'm thinking this equation is my best bet, yf = tan(angle)xf - (g/(2 * vi^2 * cos^2(angle))) * (xf^2)

    3. The attempt at a solution
    since I know the vi, xf, and yf, I was able to plug everything in and reached 795 = 1900tan(angle) - 17.375/cos^2(angle). This is where I am stuck. I tried several times but could not find a way to solve for angle.

    I would really appreciate it if someone could point out how to solve for the angle in my problem. Or is there a totally different way of solving this problem?
  2. jcsd
  3. Oct 14, 2008 #2
    Your work DOES look great (I checked your first equation). The problem is that you've got an algebraically difficult thing to solve for (the angle) because it's inside trig functions.

    What I suggest is keeping your equation in terms of the components. Relate one of these components ONLY to the initial speed and the remaining component. Then solve for the component you kept. You can use this solution to find the angle.

    Added at a later edit: You could also just graph your second function and look for where it crosses zero, but that's easy only because of modern graphing tools.
    Last edited: Oct 14, 2008
  4. Oct 14, 2008 #3
    thank you for your input, i separated my problem into x and y components:
    x component ---> 1900 = 1009cos(theta) * t
    y component ---> 795 = 1009sin(theta) * t - 4.9t^2

    I used my x component to solve for cosine
    1900/1009 = cos(theta) * t
    cos(theta) = 1.883/t

    I then used trig identity sin^2(theta) + cos^2(theta) = 1
    sin^2(theta) + (1.883/t)^2 = 1
    sin(theta) = square root(1 - (1.883/t)^2)

    then I plugged sin(theta) equation into my y component
    795 = 1009 * square root(1-(1.883/t)^2) * t -4.9t^2

    I think I'm on the right track but I don't really know where to go with this, it's not really a quadratic equation and I don't know what to do now. Any pointer would be greatly appreciated.
  5. Oct 14, 2008 #4
    nm, I got it. Thank you for all your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook