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:
\sqrt{(2g(h+\frac{R^2}{16h}))} [\tex]<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> R=\frac{v_0^2 sin(2\theta)}{g}[\tex]&lt;br /&gt; h=\frac{v_0^2 sin^2(\theta)}{2g}[\tex]&amp;lt;br /&amp;gt; v_y = v_0-g t [\tex]&amp;amp;lt;br /&amp;amp;gt; x=v_0 cos(\theta)\times t [\tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; y=v_0 sin(\theta) t-\frac {1}{2}gt^2[\tex]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; v^2=v_0^2-2gy[\tex]&amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;lt;h2&amp;amp;amp;amp;amp;gt;The Attempt at a Solution&amp;amp;amp;amp;amp;lt;/h2&amp;amp;amp;amp;amp;gt;&amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;gt; 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&amp;amp;amp;amp;amp;amp;#039;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:&amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;gt; V_0^2-2gh=\frac {2gR^2}{16h} [\tex] &amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; but this got me nowhere, then i tried to combine the other equations but this attempt was also futile.&amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; I think that i need to remove the trigonometric functions somehow but i don&amp;amp;amp;amp;amp;amp;amp;#039;t know how, and I&amp;amp;amp;amp;amp;amp;amp;#039;m hoping for some hint or something to help me move toward solving this problem.

 

[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