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Projectile Up Incline Proof

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired up an incline (incline angle φ) with an initial speed vi at an angle θi with respect to the horizontal (θi > φ). (a.) Show that the projectile travels a distance d up the incline, where

    d = 2*vi^2*cosθi*sin(θi-φ) / g*cos(φ)^2

    2. Relevant equations
    [tex]v = v_0 + a t[/tex]

    3. The attempt at a solution

    I have tried 2 different approaches to this proof. The first strategy was to tilt the figure so that the distance d was parallel to the x axis. However, through some research I found that to rotate the problem, you will also need to find the components of gravity and things got hairy from there.

    The second approach I took was to write the normal equations for x and y

    ( x= v_i cos (\theta_i) t, y= v_i sin ( \theta_i) t - 1/2 g t^2 ).

    Then I plugged x= d cos (\phi) and y = d sin ( \phi) into the 2 above equations

    I ended up with d(cos(phi)) = v_i(cos(θi))t
    and d(sin(phi)) = v_i(sin(θi)t - .5gt^2

    I was then told to isolate the t value from the first equation, and plug it into the second equation.

    However, this resulted in a very complex equation and I was unable to solve it for d, which should have given me the initial equation I was trying to prove.

    Am I making any errors, or should I try a different method?

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 10, 2007 #2


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    your second appraoch seems fine. when you isolate t from this equation:

    d(cos(phi)) = v_i(cos(θi))t

    and substitute it into the second... you should be able to solve for d... a d cancels on both sides of the equation... so it should be simple to solve for d...
  4. Sep 10, 2007 #3
    i see what you mean, but i am having trouble manipulating the variables to isolate d. so far i have:

    d(sin(phi)) =(visin(θi)vicos(θi))/dcos(phi)-(gvi^2cos^2(θi))/2d^2cos^2(θ))
  5. Sep 10, 2007 #4


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    how are you getting d in the denominator? t = dcos(phi)/vicos(θi)
  6. Sep 10, 2007 #5
    oh wow, you're absolutely right
    let me re-try this
  7. Sep 10, 2007 #6
    my second attempt definitely did not work. somehow i lost all of my "v"s

    I started with:
    dsin(phi) = vsin(θ)dcos(phi)/vcos(θ) - .5g((dcos(phi)/vcos(θ))^2

    and ended up with:
    d = (sin(phi) - tan(θ)cos(phi)) / (-gcos^2(phi))

    should i factor out the dcos(phi)/vcos(θ) initially?
  8. Sep 10, 2007 #7


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    the v's cancel in your first term on the right hand side... there's a v^2 in the denominator in the second term on the right hand side.
  9. Sep 10, 2007 #8
    ok, i'm pretty sure i've done all the math correctly, but i've gotten stuck again

    i've gotten d by itself:

    d = [2v^2cos^2(θ)sin(phi)/-gcos^2(phi)] - [2v^2cos^2(θ)sin(θ)cos(phi)/-gcos^2(phi)cos(θ)]

    i am unsure how to get this into the original form, or if i made a mistake along the way
  10. Sep 10, 2007 #9


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    use sin(A-B) = sinAcosB - cosAsinB
  11. Sep 10, 2007 #10
    it took a lot of reviewing to find all the little errors i made, but i finally got it to work out
    thank you very much for all of your help.

    im glad you gave me that trig identity, ive honestly never learned any of them. perhaps in precalc last year, but ive never used any.

    thanks again
  12. Sep 24, 2007 #11
    Hi, I hope it's ok that I am brining this thread back but I am having some problems with this proof.

    I started with the same equations he has

    and I reduce it to

    Which is not what the original poster got. As you can see my terms on the right are reversed and I have a [tex]cos^2 \theta[/tex] in one and he doesn't.

    I have gone through the math quite a few times and I can't see an error.

    Could anyone think of a possible error that I could be having? I will post my whole work if needed but it is quite long.
    Last edited: Sep 24, 2007
  13. Sep 24, 2007 #12


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    Looks the same to me, comparing with post #8.
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