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Derivitives Help!

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired with initial speed Vo at an angle Θ above the horizontal over flat ground.

    a. show that the distance d that the projectile goes and height h reached by the projectile are given by:

    d = (Vo^2 sin2Θ)/g
    h = (Vo^2 sin^2Θ)/2g

    2. Relevant equations

    X = VT

    Vx = V(cosΘ)
    Vy = V(sinΘ)

    T = d / Vo(sinΘ)

    3. The attempt at a solution

    d = (Vo^2 sin2Θ)/g
    Vt = (Vo^2 sin2Θ)/g
    (Vocos(Θ))(d / Vosin(Θ)) = (Vo^2 sin2Θ)/g

    dcot(Θ) = (Vo^2 sin(2Θ)/g
    dcot(Θ) = (Vo^2 * 2sin(Θ)cos(Θ))/g

    Then this is where I get lost.

    If I do divide cotangent of theta then would distance be proved.
    I haven't started on height yet.
  2. jcsd
  3. Sep 29, 2011 #2
    The last equation is not correct.
  4. Sep 29, 2011 #3


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    The only acceleration is -g in the y direction, the initial position is (0,0), and the initial velocity is [itex]v_x= V_0 cos(\theta)[/itex], [itex]v_y= V_0 sin(\theta)[/itex]:

    [itex]a_x= 0[/itex], [itex]a_y= -g[/itex]

    [itex]v_x= V_0 cos(\theta)[/itex], [itex]v_y= -gt+ V_0 sin(\theta)[/itex]

    [itex]x= V_0 cos(\theta)t+ 0[/itex], [itex] -(g/2)t^2+ V_0 sin(\theta)+ 0[/itex].

    Since the problem asks for the x and y distances separately, I see no reason to combine x and y and so no reason to look at [itex]tan(\theta)[/itex] or [itex]cot(\theta)[/itex]. The projectile is NOT moving in a straight line so there is no need to look for a "slope".

    (y will be largest when [itex]v_y= 0[/itex]. x will be largest when y= 0 again.)
  5. Sep 29, 2011 #4
    Yes. Think about it like this: What is the slope of the parabola at its maximum?

    It appears as though you are just trying to throw equations together, and work backwards, to get the answer. Take a minute and think about the problem.
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