Finding Max Range With 2 Angles

[Grunstadt]

Homework Statement

Today my teacher asked me "What are the best 2 angles to use when wanting to get maximum horizontal distance while on an elevated surface of 50m."

As in Dx. What would these 2 angles have to be for the projectile to go the furthest distance horizontally and what would the angles have to be for the projectile to have the highest vertical height.

How could I figure out what the 2 angles are? Is there a certain formula you would use and just plug it in?

What would X and Y have to be.

http://i.imgur.com/YLEHRzT.png?1

Homework Equations

No clue since this wasn't homework and my teacher just asked me if i could try and figure out how to solve it.
dR/d∅ = V²/g [cos∅ (cos∅ + (sin∅ * cos∅) / √sin∅² + C ) - sin∅(sin∅ + √sin∅ + C))

The Attempt at a Solution

I tried finding different equations, but this is the only one I could find and it didn't work.
I know one angle can't be above 45 since both angles have to total to 90.

 

[Mentz114]

If a projectile is fired at an angle θ to the horizintal with velocity v, then the vertical component is v sin(θ) and the horizontal is v cos(θ). The distance traveled is always v T cos(θ) where T is time spent in the air. The time spent in the air can be found like this. The vertical velocity is

V(t) = v sin(θ) -g t

and this is zero when t = v sin(θ) / g which is the at the highest point. The vertical distance is

D = H + v t sin(θ) - (1/2)g t² where H is the initial heght. So at the highest point

D = H + v sin(θ) v sin(θ) / g - (1/2)g ( v sin(θ) / g )² = H + (1/2)v² sin(θ)²/g.

The time in the air is

T² = 2D/g = (gH + v² sin(θ)²)/g²

and the square of the horizontal distance is v²cos(θ)²[(gH + v² sin(θ)²)/g²]

With H = 0 it is obvious that θ = 45^o maximises the distance. But otherwise there's no easy answer.

I found the general answer by differentiating the horizontal distance and finding the value of θ that zeros it.

cos(2θ) = g H / v²

and when H=0, the maximum is θ = 45^o which agrees.

The angle at landing α is given by tan(α) = (vertical velocity)/v cos(θ)= √(gH + v² sin(θ)²))/(vcos(θ))

and again if H = 0 the angles are equal.

[Edit : corrected an error in the last expression]