Physics homework help?

[Kai J.]

I am supposed to find that when a shell is fired with initial velocity V, and at the angle that give the maximum range, the maximum height attained is one fourth the maximum range.

I don't even know where to begin with this problem. A push in the right direction would be deeply appreciated!

 

[Jegannathan]

Yes right push with true love!
For maximum range the angle of projection 'Theta' has to be 45 degree with the horizontal.
Maximum height attained is got by the formula
Hmax = u^2 (sin45)^2 / 2g = u^2 / 4g (since sin 45 = 1 upon root 2)
Now range is given by the formula,
Rmax = u^2 sin 2x45 / g = u^2 / g (since sin 90 = 1 )
Comparing, Hmax = 1/4 Rmax. Hence proved.