Question

A stone is thrown from the top of a slope that makes an angle of 30 degree with the
horizontal. At what angle to the horizontal should the stone be thrown to get a
maximum range? Show your work step by step.

Answer 1

The initial vertical velocity is v * sin(theta) where theta = angle with the horizontal.
The horizontal velocity is v*cos(theta).

So y(t) = v*sin(theta)*t - (1/2) gt^2. The minus sign is because v is upward and g is downward.
x(t) = v*cos(theta)*t

ds/dt = 1/2[ x^2 + y^2 ] ^(-1/2) * (2x dx/dt + 2y dy/dt)

The term (x^2 + y^2), being a sum of squares, is always positive. So you don't have to worry about that term. The only thing that could make this derivative negative is the (2x dx/dt + 2y dy/dt). So your condition is

2x dx/dt + 2y dy/dt > 0
-2( v cos(theta) t) * v cos(theta) + 2* [v sin(theta) t - (1/2) g t^2] * [ v sin(theta) - g t ] > 0