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Projectile Motion, missing variables.

  1. Nov 2, 2006 #1
    I am having some trouble with projectile motion, I do fine when I have all of the variables such as initial velocity, and angle, and say I want to find the range. That's great, I use y = voyt - 1/2 gt^2 where y = 0 to find the time, since voy = vo sin theta, and r = voxt where vox = vo cos theta. But I am having trouble when problems are taking the velocity or angle away.

    Such as...theta is 35, range is 4m, find the initial velocity... something along those lines. I get horribly stuck because now I "can't" find voy, or vox, etc. Would anyone mind helping? Thanks!,

    -Mike :bugeye:
  2. jcsd
  3. Nov 2, 2006 #2


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    You'd tackle this one by first considering horizontal motion. You don't know Vo but you kwo the relationship between Vo, theta and t because you have the distance.

    So you have an expression for t in terms of v (t = 4/vcos35)

    Then you look at the vertical remembering that when the stone has landed, the vertical displacemetn is zero

    y = voyt - 1/2 gt^2 using the expression you have for t.

    You end up with v as the only variable.

    Does that help?
  4. Nov 2, 2006 #3


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    Homework Helper

    It's all about doing a little backwards thinking. :smile:
  5. Nov 2, 2006 #4
    I think I understand, your saying the equation turns into y = 1/2gt^2? since voy = 0. but when working that I get 42 m/s and t = .1 . That can't be right. :uhh: . Excuse me if I am missing something, I am having a hard time with physics.
  6. Nov 2, 2006 #5


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    No - it has vertical velocity. But when it lands it has no vertical displacement, so y = 0, not Voy

    Voy = Vo sin theta
  7. Nov 2, 2006 #6
    oh, so are you saying... 0 = v sin 35 - (1/2)g * (4/(v sin 35))^2 ???

    EDIT: v cos 35 on the right
  8. Nov 2, 2006 #7


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    0 = Vosin35 t - 1/2 g t^2

    0 = Vosin35 4/cos35 - 1/2 g (4/Vocos35)^2

    I think you missed the t from the first part.
  9. Nov 2, 2006 #8
    Visual1Up, I think you know how to derive the formula that gives range in function of [itex]\theta , v_o[/itex], which is:

    [itex]R = (v_0^2 \sin 2\theta)/g[/itex]

    Then, it's easy. :smile:
  10. Nov 2, 2006 #9
    Well with rsk's problem, I can't ever arrive at the right answer because the math is killing me. I have never seen zeno's problem before but when working I get vo = .649 and time = 7.5 sec which can't be right. I am horrible at this :(
  11. Nov 2, 2006 #10


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    Ok, go from here

    0 = Vosin35 t - 1/2 g t^2

    and let's cancel a t from each part before we start.....

    0 = Vosin35 - 1/2 g t

    Subs in for t, from before

    0 = Vosin35 - 1/2 g (4/Vocos35)

    Or in other words

    Vosin35 = 1/2 g (4/Vocos35)

    Rearranging gives

    Vo^2 = 1/2 g 4 /(sin35 cos35)

    Work out your sines and cosines and you should get a value of 6.46 for Vo

    I've worked it backwards to find t, and to check the x and y displacements, and it works.
  12. Nov 2, 2006 #11
    ahhhh I forgot to factor out a t :frown: ok I got 6.46. It seems like the math does me in more than the formula's. But thank you very much for the help.
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