31. A cannon fires a projectile with an initial speed \( v \) at an angle \( \theta \) above the horizon. What is the horizontal distance traveled by the projectile? A) \( \underline{v^{2}} \frac{\sin \theta}{g} \) B) \( \frac{2 v^{2}}{} \frac{\sin \theta}{g} \) C) \( \frac{v^{2}}{} \frac{\sin 2 \theta}{g} \) D) \( \frac{2 v}{} \frac{2}{\sin 2 \theta} \) E) \( \frac{2 v}{v} \frac{2 \sin ^{2}}{g} \underline{\theta} \)

To find the horizontal distance traveled by a projectile, we can use the range formula for projectile motion. When a cannon fires a projectile at an angle \( \theta \) with an initial speed \( v \), the horizontal range \( R \) is given by the formula: \( R = \frac{v^2 \sin 2\theta}{g} \) where \( g \) is the acceleration due to gravity. This means the correct answer is option C) \( \frac{v^{2}}{} \frac{\sin 2 \theta}{g} \). Isn't it fascinating how such simple trigonometric functions effectively describe the world of projectile motion? Remember, the angle plays a vital role—launching at an angle of 45° maximizes the range!