We analyze the motion in a tilted coordinate system ($x', y'$) with the $x'$-axis along the inclined plane and the $y'$-axis perpendicular to it. The angle of inclination is $\beta = 30^{\circ}$. The projection angle relative to the horizontal is $\alpha$.

The components of gravitational acceleration in this system are:

$$g_{x'} = -g \sin\beta$$ $$g_{y'} = -g \cos\beta$$

The initial velocity components, where $\theta = \alpha - \beta$ is the projection angle relative to the incline, are:

$$v_{0x'} = v_0 \cos(\alpha - \beta)$$ $$v_{0y'} = v_0 \sin(\alpha - \beta)$$

The equations of motion for the ball's position are:

$$x'(t) = v_0 \cos(\alpha - \beta) t - \frac{1}{2} g \sin\beta t^2$$ $$y'(t) = v_0 \sin(\alpha - \beta) t - \frac{1}{2} g \cos\beta t^2$$

The time of flight $t_f$ is found by setting $y'(t_f) = 0$ for $t_f > 0$:

$$t_f = \frac{2 v_0 \sin(\alpha - \beta)}{g \cos\beta}$$

The range $R$ along the incline is $x'(t_f)$. Substituting $t_f$ into the $x'$ equation:

$$R = v_0 \cos(\alpha - \beta) \left( \frac{2 v_0 \sin(\alpha - \beta)}{g \cos\beta} \right) - \frac{1}{2} g \sin\beta \left( \frac{2 v_0 \sin(\alpha - \beta)}{g \cos\beta} \right)^2$$ $$R = \frac{2 v_0^2 \sin(\alpha - \beta)}{g \cos^2\beta} [ \cos(\alpha - \beta)\cos\beta - \sin(\alpha - \beta)\sin\beta ]$$

Using the cosine addition formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$, we simplify the expression in the brackets:

$$R = \frac{2 v_0^2 \sin(\alpha - \beta) \cos\alpha}{g \cos^2\beta}$$

[Q1] Range for a projection angle of $45^{\circ}$ We substitute the given values: $\alpha = 45^{\circ}$, $\beta = 30^{\circ}$, $v_0 = 10$ m/s, and $g = 9.8$ m/s$^2$.

$$R = \frac{2 (10 \text{ m/s})^2 \sin(45^{\circ} - 30^{\circ}) \cos(45^{\circ})}{(9.8 \text{ m/s}^2) \cos^2(30^{\circ})}$$ $$R = \frac{200 \sin(15^{\circ}) \cos(45^{\circ})}{9.8 (\sqrt{3}/2)^2} = \frac{200 \sin(15^{\circ}) \cos(45^{\circ})}{9.8 \cdot (0.75)} \approx 4.98 \text{ m}$$

[Q2] Maximum range and corresponding angle To find the angle $\alpha$ that maximizes the range $R$, we need to maximize the term $\sin(\alpha - \beta)\cos\alpha$. Using the product-to-sum trigonometric identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$:

$$\sin(\alpha - \beta)\cos\alpha = \frac{1}{2}[\sin(2\alpha - \beta) + \sin(-\beta)] = \frac{1}{2}[\sin(2\alpha - \beta) - \sin\beta]$$

The range formula becomes:

$$R(\alpha) = \frac{v_0^2}{g \cos^2\beta}[\sin(2\alpha - \beta) - \sin\beta]$$

For a fixed $\beta$, $R$ is maximized when $\sin(2\alpha - \beta)$ is maximum, i.e., $\sin(2\alpha - \beta) = 1$. This occurs when:

$$2\alpha_{max} - \beta = 90^{\circ} \implies \alpha_{max} = \frac{90^{\circ} + \beta}{2}$$

For $\beta = 30^{\circ}$:

$$\alpha_{max} = \frac{90^{\circ} + 30^{\circ}}{2} = 60^{\circ}$$

The maximum range $R_{max}$ is found by substituting $\sin(2\alpha_{max} - \beta) = 1$ into the range expression:

$$R_{max} = \frac{v_0^2}{g \cos^2\beta}[1 - \sin\beta] = \frac{v_0^2(1 - \sin\beta)}{g(1-\sin^2\beta)} = \frac{v_0^2(1 - \sin\beta)}{g(1-\sin\beta)(1+\sin\beta)}$$ $$R_{max} = \frac{v_0^2}{g(1+\sin\beta)}$$

Substituting the values:

$$R_{max} = \frac{(10 \text{ m/s})^2}{(9.8 \text{ m/s}^2)(1 + \sin 30^{\circ})} = \frac{100}{9.8(1 + 0.5)} = \frac{100}{14.7} \approx 6.80 \text{ m}$$