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Prove this equation of initial velocity in projectile motion

  1. Apr 25, 2014 #1
    Hi every one

    1. The problem statement, all variables and given/known data
    If (R) is to be the horizontal range of a projectile and (h) its maximum height prove that the initial velocity is:
    [tex] \sqrt{(2g(h+\frac{R^2}{16h}))} [\tex]


    2. Relevant equations

    [tex]R=\frac{v_0^2 sin(2\theta)}{g}[\tex]
    [tex]h=\frac{v_0^2 sin^2(\theta)}{2g}[\tex]
    [tex]v_y = v_0-g t [\tex]
    [tex]x=v_0 cos(\theta)\times t [\tex]
    [tex]y=v_0 sin(\theta) t-\frac {1}{2}gt^2[\tex]
    [tex]v^2=v_0^2-2gy[\tex]
    3. The attempt at a solution
    I tried to put the equations of (R and h) in the given equation for initial velocity,and it did work but the problem is i don't think this method is acceptable in exams. And then I tired to start at the end by squaring both sides of the initial velocity equation and then moving ( 2gh) to the other side of the equation, which gives this:
    [tex] V_0^2-2gh=\frac {2gR^2}{16h} [\tex]
    but this got me nowhere, then i tried to combine the other equations but this attempt was also futile.

    I think that i need to remove the trigonometric functions somehow but i don't know how, and i'm hoping for some hint or something to help me move toward solving this problem.
     
  2. jcsd
  3. Apr 25, 2014 #2
    use /tex instead of \tex.
     
  4. Apr 25, 2014 #3

    haruspex

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    Allow me...
     
  5. Apr 25, 2014 #4

    haruspex

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    I agree that your R and h equations are probably not acceptable starting points.
    Your general SUVAT equations are:
    [itex]x=v_0 cos(\theta) t [/itex]
    [itex]y=v_0 sin(\theta) t-\frac {1}{2}gt^2[/itex]
    You don't care about time, so eliminate t to arrive at one equation.
    Now consider the co-ordinates at two places: max height and landing. What are x and y at those points? What equations do they give you?
     
  6. Apr 25, 2014 #5
    Thank you both
    I figured it out, what i had to do was to use the components of [tex] v_0 [/tex] to find it's magnitude:
    [tex] v_0^2=v_0^2sin^2(\theta)+v_0^2cos^2(\theta) [/tex]
    from the equation of h it is clear that: [tex] v_0^2sin^2(\theta)=2gh [/tex]
    Next i replaced [tex] sin(2\theta) [/tex] in R equation by [tex] 2cos(\theta)sin(\theta) [/tex] then i squared it and replaced [tex] sin^2(\theta) [/tex] by [tex] \frac{2gh}{v_0^2} [/tex]
    and then i put it all in place of [tex] v_0^2cos^2(\theta) [/tex] and the equation was proved.
    Thanks again
     
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