Ground to Ground Projectile: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Ground to Ground Projectile. Understanding the underlying formula is key to arriving at the correct answer below:

Range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For \(\theta = 15^\circ\), \(R_1 = \frac{u^2 \sin(30^\circ)}{g} = 1.5\text{ km}\) which implies \(\frac{u^2}{g} = 3.0\text{ km}\). For \(\theta = 45^\circ\), \(R_2 = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} = 3\text{ km}\).