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Homework Help: I'm trying to find max height of a projectile

  1. Mar 29, 2004 #1
    I'm trying to find max height of a projectile and I only know the angle of projection and the time it is in the air. No clue where to start working this problem...don't I need initial velocity to start?
  2. jcsd
  3. Mar 29, 2004 #2
    Welcome to PF. Here we will provide u with clues/hints if u show ur try.

    At max height there will be no component of velocity in y direction so Vy=0
    Eqn of trajectory:
    [tex] y=xtan(\theta) - \frac{gx^2}{2V_0^2 cos^2(\theta )}[/tex]

    Diff above to get y_max

    [tex] y=v_0 sin(\theta) t- 0.5 gt^2 [/tex]
    for [tex] y_{max} : t= \frac{v_0 sin\theta }{g} = 0.5 T [/tex]

    Where T is the time in Air
    u wont need initial velocity
    [tex]T= 2v_0 sin\theta * g^{-1} [/tex]
    so basically u know initial velocity as well
    Last edited: Mar 29, 2004
  4. Mar 29, 2004 #3
    I still don't see how I don't need the initial velocity. I can plug everything else into the equation. I'm using the equation y=v*sin(angle)*t +.5*g*t
    Is this the right equation? If so, why don't I need the velocity, and how do I get around it?
  5. Mar 29, 2004 #4
    negative max height?

    I used the equations and found the initial velocity, and it was negative, so then when I put everything back into the y-max equation I got a negative number for the y-max. I don't think that's right, where did I go wrong? Please help ASAP b/c I have class in 45 min!
  6. Mar 29, 2004 #5
    there should be -g coz velocity is in opposite direction
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