Projectile lovers (The answer is really short)

R^{2}=h^{2}+\frac{2v^{2}[sin\theta \sqrt{R^{2}-h^{2}}+hcos\theta ]cos\theta }{g}

i got something like dis

it is the same thing to wat i hav written in #11

also see http://www.targetiit.com/iit-jee-forum/posts/mechanics-doubt-17302.html [3] [3] .... same result

thnx satyapriya...nice question

There is so much literature on projectiles..If you guys are interested I can post some more..

finally i am getting....
\large required\,angle=\frac{\Pi }{4}-\frac{1}{2}\sin^{-1} (\frac{gh}{gh+v_{0}^2})

so using above relation we can find angle \phi in terms of h and u.....
then we can apply the condition for maximum range on inclined plane......and find the required angle ..............

\large R_{max}=\frac{v_{0}^2}{g(1-sin\phi ) }

write eq. for range down an inclined plane nd then find its max. value.....u will get the above equation.

don't worry about height h given in the problem......u can easily relate 'h' with the above eq. by trigonometric relations.....

i.e in the given problem we can write...
\large \frac{h}{sin\phi }= R_{max}

N.B→ in the above eq. \phi in the inclination of the imaginary inclined plane....

this sum can be solved easily by assuming an imaginary inclined plane nd using the condition for maximum range down an inclined plane.........