Jack and Jill on a Hill

So, first it should be clear that we need to find the maximum range possible for the projectile along the inclined plane. To do that, we can easily derive the formula for the projectile's range and differentiate it to get a maxima. Ok then, let's begin!

Let's consider the initial velocity of the projectile to be ' $u$ ' m/s and it's angle of projection to the inclined plane as $\alpha$ as shown in the figure. Let the angle that the inclined plane makes with the horizontal is $\theta$ which is ${30}^{\circ}$ in our case.

Notice the components!

Now, as it can be seen from the figure, the component of particle's velocity, perpendicular to the plane is given by ${ v }_{ \bot } = v\sin{\alpha}$ and that along the plane is given by ${ v }_{ \parallel } = v\cos{\alpha}$ . The perpendicular component is provided a retardation towards the plane given by ${ v }_{ \bot } = g\cos{\theta}$ whereas the parallel component is provided an acceleration along the plane given by ${ v }_{ \parallel } = g\sin{\theta}$ .

Now, the time of flight can be calculated by the double the time it takes for the perpendicular component to be zero, i.e. $T = \frac {2u\sin{\alpha}} {g\cos{\theta}}$

This is the same time that the particle will travel along the plane and it's range, $R$ can be calculated using the second equation of motion, or:

$R=(u\cos { \alpha } )\frac { 2u\sin { \alpha } }{ g\cos { \theta } } +\frac { 1 }{ 2 } (g\sin { \theta } )({ \frac { 2u\sin { \alpha } }{ g\cos { \theta } } })^{ 2 }$

This equation can be simplified to get something like:

$R=\frac { 2{ u }^{ 2 }\sin { \alpha } }{ g\cos { \theta } } (\cos { \alpha } +\frac { \sin { \alpha } \sin { \theta } }{ \cos { \theta } } )$

and then:

$R=\frac { 2{ u }^{ 2 }\sin { \alpha } \cos { (\alpha -\theta ) } }{ g\cos ^{ 2 }{ \theta } } \\$

Now, to maximise this range, we simply differentiate it with respect to $\alpha$ and equate the derivative with $0$ . Differentiating and equating, we get:

$\tan { \alpha } \tan { (\alpha -\theta ) } =0$

Putting $\tan { (\alpha -\theta ) } =\frac { \tan { \alpha } -\tan { \theta } }{ 1+\tan { a } \tan { \theta } }$ and solving the quadratic equation formed in $\tan {\alpha}$ , we get $\alpha = {150}^{\circ}$ or $\alpha = {60}^{\circ}$ . Now, since $\alpha$ can't be obtuse, hence, $\alpha = {60}^{\circ}$ is the final answer.