Here is the question

A projectile with a speed of 40m/s at an angle of 60 DEGREES above the horizontal. Find the maximum height reached by the projectile during it's flight by using conservation of energy.

Now when I go to use conservation of mechanical energy, I don't understand what to do with the final kinetic energy or how to cancel out all the masses. Please help!

the key lies in the fact that when projectile reaches max height, its entire velocity is in horizontal direction. (and vertical component is zero)
velocity in horizontal direction is always v cos theta.
now mech energy conservation
1/2 m v^2(mech energy during launch)=1/2 m (v cos theta)^2 +mg H(max)(mech energy at Hmax)
solve for H MAX. u will get v^2 sin^2 theta/2g as required.

Ricky,

That was truely a beautiful post, in fact the best post in my opinion you have ever posted!