Prove this equation of initial velocity in projectile motion

[rawezh]

Hi every one

Homework Statement

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]<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> [tex]R=\frac{v_0^2 sin(2\theta)}{g}[\tex]<br /> [tex]h=\frac{v_0^2 sin^2(\theta)}{2g}[\tex]<br /> [tex]v_y = v_0-g t [\tex]<br /> [tex]x=v_0 cos(\theta)\times t [\tex]<br /> [tex]y=v_0 sin(\theta) t-\frac {1}{2}gt^2[\tex]<br /> [tex]v^2=v_0^2-2gy[\tex]<br /> <h2>The Attempt at a Solution</h2><br /> 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:<br /> $$V_0^2-2gh=\frac {2gR^2}{16h} [\tex] <br /> but this got me nowhere, then i tried to combine the other equations but this attempt was also futile.<br /> <br /> I think that i need to remove the trigonometric functions somehow but i don't know how, and I'm hoping for some hint or something to help me move toward solving this problem.$$[/tex][/tex][/tex][/tex][/tex][/tex][/tex]

 

[dauto]

use /tex instead of \tex.

 

[haruspex]

Allow me...

Hi every one

Homework Statement

If (R) is to be the horizontal range of a projectile and (h) its maximum height prove that the initial velocity is:
$$\sqrt{(2g(h+\frac{R^2}{16h}))}$$

Homework Equations

$$R=\frac{v_0^2 sin(2\theta)}{g}$$
$$h=\frac{v_0^2 sin^2(\theta)}{2g}$$
$$v_y = v_0-g t$$
$$x=v_0 cos(\theta)\times t$$
$$y=v_0 sin(\theta) t-\frac {1}{2}gt^2$$
$$v^2=v_0^2-2gy$$

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:
$$V_0^2-2gh=\frac {2gR^2}{16h}$$
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.

 

[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?

 

[rawezh]

Thank you both
I figured it out, what i had to do was to use the components of $$v_0$$ 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