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:

Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).