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