1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating angle for trajectory with exit position not at 0,0

  1. Jan 9, 2010 #1

    xal

    User Avatar

    I'd like to calculate the angle to shoot a projectile from a cannon in order to hit a target. However, the position that the projectile will be exiting from will not be at 0, 0 -- it will vary according to the angle of the cannon itself. The projectile will be exiting from the tip of the cannon, but the cannon will be rotated about a different point (both points, as well as the target position, will be on the same plane though).

    If the projectile was exiting from 0, 0, then I could use this:
    [tex]\tan \left( \theta \right) ={\frac {{{\it v0}}^{2}+\mbox {{\tt `\&+-`}
    } \left( \sqrt {{{\it v0}}^{4}-{g}^{2}{x}^{2}-2\,gy{{\it v0}}^{2}}
    \right) }{gx}}
    [/tex]

    However, in this case, I was considering replacing [tex]x[/tex] and [tex]y[/tex] with [tex]x-l\cos \left( \theta \right)[/tex] and [tex]y-l\sin \left( \theta \right) [/tex] respectively, where [tex]l[/tex] would be the distance away the tip.

    That results in this:
    [tex]\tan \left( \theta \right) ={\frac {{{\it v0}}^{2}+\sqrt {{{\it v0}}^{
    4}-g \left( g \left( x-l\cos \left( \theta \right) \right) ^{2}+2\,
    \left( y-l\sin \left( \theta \right) \right) {{\it v0}}^{2} \right)
    }}{g \left( x-l\cos \left( \theta \right) \right) }}


    [/tex]

    Different form:
    [tex]-1/2\,{\frac {g \left( x-l\cos \left( \theta \right) \right) ^{2}
    \left( \tan \left( \theta \right) \right) ^{2}}{{v}^{2}}}+x\tan
    \left( \theta \right) -1/2\,{\frac {g \left( x-l\cos \left( \theta
    \right) \right) ^{2}}{{v}^{2}}}-y+l\sin \left( \theta \right) =0
    [/tex]

    But I am not sure if this is a way to do it, and I am not sure to go about solving for the angle either.
     
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted