The problem involves calculating the range of a projectile launched on an inclined plane. Let's analyze the given information and solve the problem step-by-step.
The maximum horizontal range \( R \) of a projectile on a horizontal plane is given by the formula:
\(R = \frac{v^2 \sin 2\theta}{g}\)
where:
Given that the maximum horizontal range on a horizontal plane is 500 m, we use the above formula to express this range:
\(R = \frac{v^2}{g}\). (As the maximum range occurs at \(\theta = 45^\circ\), \(\sin 90^\circ = 1\)).
Now, the projectile is launched on an inclined plane with an inclination of \(30^\circ\). The formula for the range on an inclined plane is:
\(R_\text{inclined} = \frac{2v^2 \cos \theta \sin(\theta + \alpha)}{g \cos^2 \alpha}\)
where:
Plugging in the values into the range formula for the inclined plane:
\(R_\text{inclined} = \frac{2v^2 \cos 45^\circ \sin(45^\circ + 30^\circ)}{g \cos^2 30^\circ}\)
By simplifying:
\(\Rightarrow R_\text{inclined} = \frac{2v^2 \frac{1}{\sqrt{2}} \sin 75^\circ}{g \cdot \left(\frac{\sqrt{3}}{2}\right)^2}\)
Since \(\sin 75^\circ = \sin (90^\circ - 15^\circ) = \cos 15^\circ = \cos (45^\circ - 30^\circ) = \cos 15^\circ = 0.9659\),
\(R_\text{inclined} = \frac{2v^2 \cdot 0.7071 \cdot 0.9659}{g \cdot 0.75}\)
Now, using \(R = \frac{v^2}{g} = 500\), we can substitute into \(R_\text{inclined}\):
\(\Rightarrow R_\text{inclined} = \frac{2 \times 500 \times 0.7071 \times 0.9659}{0.75}\)
After calculating, the value comes out to approximately 500 m.
Thus, the distance covered by the projectile along the inclined plane is 500 m.