derivation of range formula when initial height is differnt from final

**superdude** · April 18th 2010, 02:26 PM

What is the derivation for this formula?
$R=(1+\sqrt{1-\frac{2gy}{v_o^2sin^2\theta_o}})\frac{v_o^2}{2g}si n2\theta_o$
R=horizontal displacement
y=final height-initial height

**ADARSH** · April 27th 2010, 01:37 AM

Originally Posted by superdude

What is the derivation for this formula?
$R=(1+\sqrt{1-\frac{2gy}{v_o^2sin^2\theta_o}})\frac{v_o^2}{2g}si n2\theta_o$
R=horizontal displacement
y=final height-initial height

Take a look at me

Give it a healthy reading...till maximum range ...take care of difference of signs in your question and in Wikipedia's proof...If you still find any trouble feel free to ask.

Adarsh

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