3. (a) A particle is projected from a point on horizontal ground - Leaving Cert Applied Maths - Question 3 - 2013

From the given information, the maximum height $H$ reached by the particle is given by:

$H = \frac{R}{4 \sqrt{3}}$

Using the formula for maximum height:

$H = \frac{u^{2} \sin^{2} \alpha}{2g}$

Equating the two expressions for height:

$\frac{u^{2} \sin^{2} \alpha}{2g} = \frac{2u^{2} \sin \alpha \cos \alpha}{4g \sqrt{3}}$

This simplifies to:

$4u^{2} \sin^{2} \alpha = 4u^{2} \sin \alpha \cos \alpha \sqrt{3}$

Dividing both sides by $u^{2}$ and rearranging gives:

$\tan \alpha = \frac{1}{\sqrt{3}}$

Thus, we find:

$$\alpha = 30^{\circ}.$

Given that the plane is inclined at an angle $\theta_{0} = \tan^{-1} \frac{1}{2}$, we consider the equations for projectile motion up the slope. The range up the inclined plane is given by:

$R = \frac{u^{2} \sin 2\theta}{g \cos \theta_{0}}$

To maximize the range, we differentiate this expression with respect to $\theta$ and set the derivative to zero:

[{\frac{dR}{d\theta} = 0 = 2u^{2} \left(\cos 2\theta \cdot \frac{1}{g \cos \theta_{0}}\right) - \frac{u^{2} \sin 2\theta \cdot \sin \theta_{0}}{g \cos^2 \theta_{0}}}]

Solving the resulting equation yields:

$2\cos 2\theta - \sin \theta_{0} = 0$

Plugging in $\theta_{0}$:

$2\cos 2\theta - 2 = 0$

This simplifies to find:

$\theta = 31.7^{\circ}$ as the angle for maximum range.