A particle is projected up a plane inclined at an angle $b$ to the horizontal, the angle of projection being $a$ above the horizontal.
Part 1: If the initial speed of projection is $V$ m/s show that range, $C$, on the inclined plane is given by: $$C= 2 V^2 \frac{\sin(a-b) \cos(a)}{g\cos(b)^2}$$
Part 2: Determine the maximum range ($C$ max) on the plane and the value of $b$ for which this occurs. maximising the trigonometric function is the main problem here
Here is a diagram:
On
Here is a diagram of the problem with a few helpful elements added:
The figure illustrates the fact that if not for the influence of gravity, the projectile would simply travel a distance $Vt$ along the line at an angle $a$ above the horizontal in a certain time $t,$ but because of gravity, the projectile is deflected downward a distance $\frac 12 gt^2,$ so it ends up at the intersection of the blue parabola and the vertical red line. This point is the landing point on the inclined surface if we choose $t$ to be the amount of time the projectile is in flight before striking that surface.
Now we do a little trigonometry. We draw a right triangle with the segment $Vt$ as the hypotenuse and an extension of segment $C$ as one of the legs. The vertical red line cuts a smaller right triangle from this triangle; the smaller triangle has hypotenuse $\frac 12 gt^2$ and angle $b$ at the topmost vertex, hence the adjacent leg is $\frac 12 gt^2 \cos b.$ But that leg is also the leg opposite the angle $a-b$ of the larger right triangle, so we have $$Vt \sin (a-b) = \frac 12 gt^2 \cos b.$$
Now consider the horizontal segment in the figure, which is the leg adjacent to angle $b$ in a right triangle whose hypotenuse is $C,$ so its length is $C \cos b,$ but it is also the leg adjacent to angle $a$ in a right triangle whose hypontenuse is $Vt.$ This gives us $$C \cos b = Vt \cos a.$$
From these two equations we can conclude that $$\left(\frac 12 gt^2 \cos b\right)(C \cos b) = (Vt \sin (a-b))(Vt \cos a).$$
We have already assumed that $t > 0,$ so we can cancel factors of $t.$ Assuming $b$ is less than a right angle, $\cos b > 0$ as well, and we can assume $g > 0.$ We can use these facts to isolate $C$ on one side of the equation. The result is $$C = \frac{2V^2 \sin (a-b) \cos a}{g \cos^2 b}.$$
The second part of the problem is to maximize $C.$ At least one version of the posted question allows us to vary $b$ in order to maximize $C.$ But we can make $C$ as large as we want merely by keeping $a$ constant and letting $b$ approach $-\frac\pi2$ radians--that is, incline the plane as steeply downwards as we like, but never quite vertical (because then there is no second intersection of the trajectory with the plane, and $C$ is undefined). We can even allow $a$ to vary and still $C$ is unbounded as long as $a$ is not too close to $b.$
So I will answer a more interesting question, which I suspect is what was meant to be asked: find $a$ that maximizes $C$ for given constant values of $V,\ g,$ and $b.$
To solve this, observe the trigonometric identity $$\sin x - \sin y = 2 \sin\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right).$$ Conveniently, there is a product of a sine and cosine in the quantity we want to maximize. So we would like to have $$\begin{eqnarray} a - b &=& \frac12(x-y), \\ a &=& \frac12(x+y). \end{eqnarray}$$ Solving for $x$ and $y$ in terms of $a$ and $b,$ we find that $$\begin{eqnarray} x &=& 2a - b, \\ y &=& b. \end{eqnarray}$$ Therefore, we can use the identity to substitute for $\sin (a-b) \cos a$ as follows: $$C = \frac{2V^2}{g \cos^2 b} (\sin (2a-b) - \sin b).$$
But according to the conditions of the problem, everything on the right-hand side of this equation is a constant except for $a.$ We therefore maximize $C$ by maximizing $\sin (2a-b),$ which occurs when $2a-b$ is a right angle. Measuring in radians, that gives us $$a = \frac12\left(\frac\pi2 + b\right).$$ Measured in degrees, $a = 45 + \frac12 b.$
Here's me having a go at it:
Now, let the particle hit the plane after time t.
Horizontal Displacement = $C*\cos b$
Vertical Displacement = $C*\sin b$
Horizontal velocity = $V*\cos a$ (along positive x-axis)
Vertical velocity = $V*\sin a$ (along positive y-axis)
After time t:
$V*\cos a*t = C*\cos b$ . . . (1)
$V*\sin a *t - 0.5*g*t^2 = C*\sin b$ . . . (2)
From (1) obtain the value of $t$ and plug it into (2) and rearrange to answer Qn 1