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Max height of trajectory

  1. Feb 27, 2008 #1
    1. I am trying to derive the equation for maximum height of a trajectory. Must be in terms of v, theta, and g.

    2.Vy=vsin(theta) dy=vsin(theta)t+0.5gt^2

    3. I know that I must find the point when the vertical velocity is equal to zero and must find one equation for time. Then sub this equation into another.

    So I found from the first equation that t=dysin(theta) and then substituted this equation to the second equation listed above. I then got this equation dy=vdsin(theta)+0.5g(dsin(theta))^2.
    I dont know what to do from here because of the two displacements. And I dont even know what im doing is correct as I cant solve.
    Can anyone help lead me in the direct equation, it would also be helpful to acctually show me what the max height equation looks like.
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2
    You're going to need more than one equation to figure this out, but I'll narrow it down for you. We're given that the final velocity vy is equal to zero, and we're given g, v, and theta. We want to begin by placing the initial velocity in the x-direction and the initial velocity in the y-direction in terms of v and theta. To do so we use the equations:

    [tex] v_{0x} = v cos \theta [/tex]

    [tex] v_{0y} = v sin \theta [/tex]

    From the kinematic equations, we'll want to use:

    [tex] y = v_{0y} - 0.5(g)t^2 [/tex]

    You know how to get v 0y in terms of v and theta, so now we focus on how to get the time t into the given terms. Note again that the final velocity in the y-direction is zero, thus:

    [tex] 0 = v_{0y} -gt [/tex]

    Solve this for t and you're in business.
     
    Last edited: Feb 27, 2008
  4. Feb 27, 2008 #3
    In the kinematics equation, what happened to the t?
    and can you explain what Voy is opposed to Vy?
     
  5. Feb 27, 2008 #4
    Oh sorry you're right, that should be:

    [tex] y = v_{0y}t - 0.5(g)t^2 [/tex]

    Also, v0y is the initial velocity in the y-direction, and vy is the final velocity in the y-direction, which should be zero, right?

    I forgot to put the "v" in the initial velocity equations the first time around. Just to clarify, they should be:

    [tex] v_{0x} = v sin \theta [/tex]

    [tex] v_{0y} = v cos \theta [/tex]

    Does this make sense?
     
  6. Feb 27, 2008 #5
    can you possibly explain how Voy-gt equals zero?
     
  7. Feb 27, 2008 #6
    One of the kinematic equations for the y-direction is [tex] v_y = v_{0y} -gt [/tex]. If we know the final velocity in the y-direction to be zero, then we have:

    [tex] 0 = v_{0y} -gt [/tex]

    Does that make sense?
     
  8. Feb 27, 2008 #7
    That would be 0=Voy-g*tfinal then

    In general 0=Voy-gt is not true, except at tfinal, which is the time when it reaches the peak
     
  9. Feb 27, 2008 #8
    ok everything makes sense now.
    i substituted the equation and came up to
    dy=(v^2cos(theta)^2)/g) - (vcos(theta)/2)
    Have i done this right so far? if so, how can i continue?
     
  10. Feb 27, 2008 #9
    That would have been right if I had got more than three hours of sleep, I do apologize. It should be sine of theta, not cosine in your equation. You'd use:

    v0x = vcos \theta

    v0y = v sin \theta

    But substitute the sin in for the cos, and you're good to go.
     
  11. Feb 27, 2008 #10
    Oops, one last edit. You're missing one term, and a few squares. Your answer should be:

    dy=(v^2sin(theta)^2)/g) - (v^2sin(theta)^2/g2)
     
  12. Feb 27, 2008 #11
    ok thank!!
     
  13. Feb 27, 2008 #12
    You're welcome, but also note that the final answer can be simplified :)
     
  14. Feb 27, 2008 #13
    ya i got it, thanks again
     
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