Question

A plane surface is inclined making an angle theta with the horizontal. From the bottom of this inclined plane, bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is

Answer 1

See the diagram.  

Let the bullet be fired at angle   ∠Ф+θ  with the horizontal.  The equations of motion of the bullet are:

x = v cos(Ф+ θ)  * t
y =  v sin(Ф+ θ)  t - 1/2 * g t²
or, the trajectory :   y = x tan (Ф+ θ) -  g  x² Sec² (Ф+θ) /(2v²)     --- (1)

At the point of maximum trajectory on the slope (intersection with parabola), we have   y = x tanθ .  So (1)  becomes

=>  x [ tan(Ф+θ) - tan θ ] = (g/2v²) x² sec² (Ф+θ )
=>  x = 0  or,
   (sin (Ф+ θ) cosθ  - cos(Ф+ θ) sin θ = (g/2v²) x sec² (Ф+θ ) cos(Ф+θ ) cosθ 
=> 2 sinФ * cos (Ф+θ ) = (g/v²) x cos θ
=> x = [ Sin(2Ф+θ ) - sinθ ] * [v² Secθ /g  ]

Range R on the slope = R = x sec θ
=> R = [ v² Sec²θ / g) [ Sin (2Ф +θ ) - Sin  θ]

Maximizing R wrt variable angle Ф, means   2Ф+ θ = π/2
=>  Ф = π/4  - θ/2

Then maximum range along the slope
 = (v² Sec²θ / g) [ 1 - sin θ]
 =  v²/ [g (1+ sinθ) ]            answer.
 or,  = [ v² * Cosec² (π/4 + θ/2) ] /(√2 * g)

Answer 2

Solution:-
Let the bullet be fired with velocity v from the inclined plane which makes the angle theta with the horizontal makes angle α with the inclined plane.
The component of initial velocity parallel and perpendicular to the plane is 
V II = V cos α
V ⊥ = V sin α
The component g is given by
a II - g sin theta
a ⊥ g cos theta
Therefore the particle decelerates at rate of g sin theta from O to P.
Time of flight of inclined plane is
T = 2V sin α/g cos theta
Maximum height of the inclined plane is
H = V² sin² α/g cos theta
The horizontal range of bullet is
In one dimensional motion s = ut + 1/2 at²
Horizontal range of inclined plane is H = V II T + 1/2 a II T²
H = V cos α(2V sin α/g cos theta) - 1/2 g sin theta(2V sin α/g cos theta)
H = 2V²/g{sinα sin(theta + α)}/cos² α
Maximum range occurs when α = (π/4) - (theta/2)
Thus maximum range bullet fired from inclined plane is 
H = V²/g(1 + sin theta)  Answer.