[Q1] Relationship between Product of Flight Times and Range

We establish a coordinate system with the x-axis along the incline and the y-axis perpendicular to it. The acceleration due to gravity $\vec{g}$ has components:

$$g_x = -g\sin\theta$$ $$g_y = -g\cos\theta$$

The initial velocity $\vec{v}_0$ at an angle $\alpha$ to the incline has components:

$$v_{0x} = v_0\cos\alpha$$ $$v_{0y} = v_0\sin\alpha$$

The equations of motion for a particle are:

$$x(t) = (v_0\cos\alpha)t - \frac{1}{2}g\sin\theta t^2$$ $$y(t) = (v_0\sin\alpha)t - \frac{1}{2}g\cos\theta t^2$$

At landing, after a time of flight $t$, the particle is at position $(R, 0)$. From the y-equation, for $t eq 0$:

$$0 = (v_0\sin\alpha)t - \frac{1}{2}g\cos\theta t^2 \implies v_0\sin\alpha = \frac{1}{2}gt\cos\theta$$

From the x-equation:

$$R = (v_0\cos\alpha)t - \frac{1}{2}g\sin\theta t^2 \implies v_0\cos\alpha = \frac{R}{t} + \frac{1}{2}gt\sin\theta$$

To eliminate the launch angle $\alpha$, we use the identity $\sin^2\alpha + \cos^2\alpha = 1$:

$$(v_0\sin\alpha)^2 + (v_0\cos\alpha)^2 = v_0^2$$ $$\left(\frac{1}{2}gt\cos\theta\right)^2 + \left(\frac{R}{t} + \frac{1}{2}gt\sin\theta\right)^2 = v_0^2$$

Expanding and simplifying the equation:

$$\frac{g^2t^2\cos^2\theta}{4} + \frac{R^2}{t^2} + R g\sin\theta + \frac{g^2t^2\sin^2\theta}{4} = v_0^2$$ $$\frac{g^2t^2}{4}(\cos^2\theta + \sin^2\theta) + \frac{R^2}{t^2} + R g\sin\theta = v_0^2$$ $$\frac{g^2t^2}{4} + \frac{R^2}{t^2} + R g\sin\theta = v_0^2$$

To solve for $t$, we can arrange this into a quadratic equation in $t^2$. Multiplying by $t^2$:

$$v_0^2 t^2 = \frac{g^2t^4}{4} + R^2 + R g\sin\theta t^2$$ $$\frac{g^2}{4}t^4 - (v_0^2 - R g\sin\theta)t^2 + R^2 = 0$$

This is a quadratic equation for $t^2$. For a given range R, there are two possible times of flight, $t_1$ and $t_2$. Their squares, $t_1^2$ and $t_2^2$, are the two roots of this equation. According to Vieta's formulas, the product of the roots is given by the ratio of the constant term to the leading coefficient:

$$t_1^2 t_2^2 = \frac{R^2}{g^2/4} = \frac{4R^2}{g^2}$$

Taking the square root (since time must be positive), we find the relationship:

$$t_1 t_2 = \frac{2R}{g}$$

[Q2] Comparison with Horizontal Ground

For projectile motion on horizontal ground, the inclination angle is $\theta=0$. The same range R is achieved for two complementary launch angles $\alpha_1$ and $\alpha_2 = \pi/2 - \alpha_1$.

The times of flight are:

$$t_1 = \frac{2v_0\sin\alpha_1}{g}$$ $$t_2 = \frac{2v_0\sin\alpha_2}{g} = \frac{2v_0\sin(\pi/2-\alpha_1)}{g} = \frac{2v_0\cos\alpha_1}{g}$$

The product of the flight times is:

$$t_1 t_2 = \left(\frac{2v_0\sin\alpha_1}{g}\right)\left(\frac{2v_0\cos\alpha_1}{g}\right) = \frac{2v_0^2(2\sin\alpha_1\cos\alpha_1)}{g^2} = \frac{2v_0^2\sin(2\alpha_1)}{g^2}$$

The range on horizontal ground is $R = \frac{v_0^2\sin(2\alpha_1)}{g}$. Substituting this into the product of times expression gives:

$$t_1 t_2 = \frac{2}{g} \left(\frac{v_0^2\sin(2\alpha_1)}{g}\right) = \frac{2R}{g}$$

The result for the product of flight times is identical for both motion on an incline and motion on horizontal ground. The relationship $t_1 t_2 = 2R/g$ is surprisingly independent of the inclination angle $\theta$.