Prove this equation of initial velocity in projectile motion

1. Apr 25, 2014

rawezh

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:
$$\sqrt{(2g(h+\frac{R^2}{16h}))} [\tex] 2. Relevant equations [tex]R=\frac{v_0^2 sin(2\theta)}{g}[\tex] [tex]h=\frac{v_0^2 sin^2(\theta)}{2g}[\tex] [tex]v_y = v_0-g t [\tex] [tex]x=v_0 cos(\theta)\times t [\tex] [tex]y=v_0 sin(\theta) t-\frac {1}{2}gt^2[\tex] [tex]v^2=v_0^2-2gy[\tex] 3. The attempt at a solution I tried to put the equations of (R and h) in the given equation for initial velocity,and it did work but the problem is i don't think this method is acceptable in exams. And then I tired to start at the end by squaring both sides of the initial velocity equation and then moving ( 2gh) to the other side of the equation, which gives this: [tex] V_0^2-2gh=\frac {2gR^2}{16h} [\tex] but this got me nowhere, then i tried to combine the other equations but this attempt was also futile. I think that i need to remove the trigonometric functions somehow but i don't know how, and i'm hoping for some hint or something to help me move toward solving this problem. 2. Apr 25, 2014 dauto use /tex instead of \tex. 3. Apr 25, 2014 haruspex Allow me... 4. Apr 25, 2014 haruspex I agree that your R and h equations are probably not acceptable starting points. Your general SUVAT equations are: $x=v_0 cos(\theta) t$ $y=v_0 sin(\theta) t-\frac {1}{2}gt^2$ You don't care about time, so eliminate t to arrive at one equation. Now consider the co-ordinates at two places: max height and landing. What are x and y at those points? What equations do they give you? 5. Apr 25, 2014 rawezh Thank you both I figured it out, what i had to do was to use the components of [tex] v_0$$ to find it's magnitude:
$$v_0^2=v_0^2sin^2(\theta)+v_0^2cos^2(\theta)$$
from the equation of h it is clear that: $$v_0^2sin^2(\theta)=2gh$$
Next i replaced $$sin(2\theta)$$ in R equation by $$2cos(\theta)sin(\theta)$$ then i squared it and replaced $$sin^2(\theta)$$ by $$\frac{2gh}{v_0^2}$$
and then i put it all in place of $$v_0^2cos^2(\theta)$$ and the equation was proved.
Thanks again