Projectile at an Angle ∅ Above Inclined Plane of Angle θ

Part A

The projectile is being fired at an angle ∅+θ above the horizontal. So the trajectory equation is

y(x) = tan(∅+θ)x - g(x^2)/(2(Subscript[v, 0]^2)(cos^2)(∅+θ))      where (Subscript[v, 0])is the speed.

Now if the projectile hits the plane at coordinates (x,y), then we know the expression for tanθ

tanθ = y/x  which gives  y = x tanθ .  We can set these two expression for y equal to each other

x tanθ = tan(∅+θ)x - g(x^2)/(2(Subscript[v, 0]^2)(cos^2)(∅+θ))

This equation has a solution at x = 0, corresponding to y = 0, meaning when the projectile was fired.

There is a second solution when x is not 0, which we can derive by first dividing both sides by x

tanθ = tan(∅+θ) - gx/(2(Subscript[v, 0]^2)(cos^2)(∅+θ))        which leads to this equation for x on the left side

x = [ tan(∅+θ) - tanθ]2(Subscript[v, 0]^2)(cos^2)(∅+θ)

Last, we can divide by cosθ to get the distance along the inclined plane (range)

x/cosθ = [ tan(∅+θ) - tanθ]2(Subscript[v, 0]^2)(cos^2)(∅+θ)/cosθ

Part B

To get the maximum range, we need to take the derivative with respect to the angle ∅ of the right hand side of the last equation, and set that derivative to 0.  There is a commercial computer program called Mathematica which can do derivatives and algebraic operations easily, although there is a learning curve of a few days time on getting the Mathematica syntax correct. We can simplify the problem by dropping the factor 2(Subscript[v, 0]^2)/cosθ which do not depend upon the angle ∅. So we are left with the task of taking the derivative of the expression [tan(∅+θ) - tanθ](cos^2)(∅+θ) .  In the following, the text in RED is an input command to Mathematica, while the text in MAGENTA is the output result of that command. First we find the derivative with respect to the angle ∅, and the Mathematica command is

D[(Tan[∅ + θ]  -  
              Tan[θ])*(((Cos[∅ + θ]))^2), ∅]

The above is the derivative result, and we can simplify it using trig identities to obtain

FullSimplify[
        1 - 2 Cos[θ + ∅] Sin[θ + ∅] (-Tan[θ] + Tan[θ + ∅])]

Lastly, we can set the above result equal to 0 in order to find the value of ∅ which gives a maximum. By simple inspection you can see that θ+2∅=π/2 is the correct answer since it makes the cosine term to be 0.

Solve[
        Cos[θ + 2 ∅] Sec[θ] == 
          0,  ∅]

The second (positive) solution  ∅ = π/4 - θ/2 is what we are looking for, since negative angles would not have any physical meaning in this problem.