**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.

http://ift.tt/1ffVgky