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Help find initial velocity

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I need help finding the x and y components of initial velocity being given the angle the trajectory was fired at and the distance the projectile traveled. For this example I'm using an angle of 15° and the projectile travels 0.534m. I'm not sure if I'm going in the right direction in trying to solve this problem.


    2. Relevant equations
    Vx = Vcos(angle)
    Vy = Vsin(angle)

    x -x initial = Vx * t


    3. The attempt at a solution
    Solved those to get: Vx = Vcos(15) and Vy = Vsin(15).

    After pluggin Vx into the equation, I'm having trouble figuring out where to go from here. Any help is appreciated.
     
  2. jcsd
  3. Oct 4, 2012 #2

    tiny-tim

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    welcome to pf!

    hi gsg822! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    use the x equation to find t (as a function of V), then use that value of t in the y equation …

    show us what you get :smile:
     
  4. Oct 4, 2012 #3
    When you say to use the y equation are you referring to y = y0 + voy * t - 0.5 * g * t2. I've tried using this equation and I get stuck when I try and simplify it down.

    I solved the x equation for time and got t = 0.534m / (V * sin(15))
     
  5. Oct 4, 2012 #4

    tiny-tim

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    yup! :smile:
    if i'm reading the question correctly, y = yo :wink:
     
  6. Oct 4, 2012 #5
    I have two variables, V and Voy.

    0 = Voy(0.534m / Vsin(15)) - (1/2)(9.8m/s2)(0.534m / Vsin(15))2

    (1/2)(9.8m/s2)(0.285m / (Vsin(15)2) = Voy(0.534m / Vsin(15))

    2.79m2/s2 / ( Vsin(15)2) = 2Voy(0.534m / Vsin(15))

    2.79m2/s2 / ( Vsin(15)) = 2Voy(0.534m)

    I get stuck here and don't know where to take it from here. I'm also not really sure if this work I did is even right. Am I heading in the right direction?
     
  7. Oct 4, 2012 #6

    tiny-tim

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    Voy = Vcos15° :wink:

    (and you could have divided the whole of that equation by t, since you had 0 on the LHS)

    and now i'm off to bed :zzz:
     
  8. Oct 5, 2012 #7
    I forgot that I could substitue Vsin(15) for Voy.

    I think I may have gotten the answer.

    Vsin(15) = 5.2 / 2Vcos(15)

    Rearranged that to V2 = 5.2 / (2sin(15)cos(15))

    V = sqrt(5.2 / (2sin(15)cos(15)))

    V = 3.22 m/s

    Plug that value back into the Vox and Voy equations to get:

    Vox = 3.11 /s
    Voy = 0.833 m/s

    Could you verify if this is correct?
     
  9. Oct 9, 2012 #8

    tiny-tim

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    hi gsg822! :smile:

    (i'm sorry for the delay :redface:)

    where did this come from? …
     
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