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Projectile Motion cannon off a cliff

  1. May 24, 2009 #1
    Okay there is an object being shot from a cannon at an angle off a cliff. You are given the distance from the bottom of the cliff, the initial velocity, and the height of the cliff. What is an equation that would use these knowns to give you the measure of the angle that the object was shot at. Can someone help me with this?
  2. jcsd
  3. May 24, 2009 #2
    think about how you would find the range of a projectile if you were given the height of the cliff and the initial velocity.

    you can then work backwards to find theta.

    start with splitting the initial velocity into orthogonal components.
  4. May 24, 2009 #3


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    Homework Helper

    These equations might be a useful resource for you:
  5. May 24, 2009 #4
    Okay, I get an equation with both sine and cosine terms that I can't get rid of. Can someone just show me how this is done?
  6. May 24, 2009 #5


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    Homework Helper

    Typically the trig identity

    2*sinθ*cosθ = sin2θ

    is of use in resolving Range equations.

    Without seeing your work, there's no useful way to help you with what you are doing that I can see.
  7. May 24, 2009 #6
    show us the equation and we'll help....

    or just remember that sin(x)/cos(x) = tan(x)

    edit: yeah or the trig identity posted above
  8. May 25, 2009 #7
    Using earlofwessex's idea, I start with R=((V0/g)(V0sinΘ+sqrt((V0sinΘ)^{2}+2gh) where V0 is initial velocity, Θ is the angle, R is range, g is acc due to gravity, and h is the height of the cliff.

    I tried to solve this equation for Θ and get:


    Am I using the wrong equation or there another trig identity or what?
  9. May 25, 2009 #8
    Towk667 said that the distance from the bottom of the cliff and the height of the cliff were given seperately. So that must mean the cannon shot the ball from a distance from the end of the cliff. Keeping this in mind, I found the horizontal range, i.e., the distance from the cannon(considering its distance from the end of the cliff) to the place where the ball hits the ground:

    {[ucosӨ * sqrt(u^2sin^2Ө + 2gh)] +[2u^2sin2Ө]}/2g = R

    I don't know what to do ahead of this, but I'm quite sure this is the correct equation.
  10. May 25, 2009 #9
    I've got another question related to projectile motion:
    "What is the average velocity of a projectile between the instants it crosses half the maximum height? It is projected with a speed 'u' at an angle Ө with the horizontal"

    (a) u sinӨ (b) u cosӨ (c) u tanӨ (d) u

    Could I get some help on this one? I don't even know where to start.
  11. May 25, 2009 #10
    I think I can help with this one. Think about the formula to find maximum height and then use that knowledge and a formula for final velocity that doesn't use time to get your answer.
  12. May 26, 2009 #11
    I did what you said and got the following equation for the final velocity:
    sqrt{ [ (2g (u sinӨ)^2) + (2g (u cosӨ)^2) - ((u sinӨ)^2)] / 2g }

    Now, what do I do? I don't think I can use the final velocity to evaluete the average velocity, right?
    Is there any other method in which I can find the total distance and total time?
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