It is well known that a ball thrown at a fixed speed along a flat field will travel farthest if it is thrown at an angle of 45 degrees. (The ball is thrown and caught at the same height, and we neglect air resistance. )

The above shows the path of a ball with initial speed 66ft/sec. thrown and caught at a height of 5 ft.
Also, I take g = -32ft/sec², but that doesn't factor into A.

Find the angle A at which the ball will travel farthest if it is thrown down a hill that is at a constant angle B with a horizontal.

The above shows the path of a ball with initial speed 66ft/sec. thrown down a hill with B = p/8 = 22.5 degrees.

Also find the angle A at which the ball will travel farthest if it is thrown up a hill that is at a constant angle of B with a horizontal.

The above shows the path of a ball with initial speed 66ft/sec. thrown up a hill with B = p/8 = 22.5 degrees.

Let V be the speed of the throw at angle A. For the uphill throw at time t, the ball is at

x(t) = Vcos(A)t, and y(t) = Vsin(A)t – ½gt². The ball hits the hill when y/x = tan(B).

This leads to Vsin(A)t – ½gt² = tan(B)Vcos(A)t, so t = 0 or t = (2V/g)[sin(A) – tan(B)cos(A)].

Plugging the later into x(t) gives x = Vcos(A)(2V/g)[sin(A) – tan(B)cos(A)]. Simplifying with double angle identities we have x = V²/g[sin(2A) – tan(B)cos(2A)-tan(B)]. To find the angle A that maximizes x, we differentiate wrt A,. This leads to tan(B) = -cot(2A), or tan(B) = -tan(π/2 – 2A) = tan(2A – π/2). Therefore B = 2A– π/2 or

A = π/4 + B/2.

For the downhill throw at time t, the ball is at x(t) = Vcos(A)t, and y(t) = Vsin(A)t – ½gt². The ball hits the hill when y/x = -tan(B). As above, this leads to tan(B) = cot(2A), or tan(B) = tan(π/2 – 2A) .

Therefore B = π/2 -2A or

A = π/4 - B/2.