A projectile is launched from ground level with an initial velocity of \( v \) at an angle \( \theta \) above the horizontal. Ignore air resistance. 13. The maximum horizontal displacement of the projectile is A) \( \frac{v \sin 2 \theta}{g} \) B) \( \frac{(v \sin \theta)^{2}}{\text { C) } \frac{v \sin 2 \theta}{2 g}} \begin{array}{l}\text { D) } \frac{(v \sin \theta)^{2}}{v g} \\ \text { E) } \frac{\nu}{2} \frac{\sin 2 \theta}{g}\end{array} \)
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To find the maximum horizontal displacement of a projectile launched at an angle \( \theta \) with initial velocity \( v \), we first determine the time of flight. The time the projectile spends in the air is given by \( T = \frac{2v \sin \theta}{g} \), where \( g \) is the acceleration due to gravity. The horizontal range \( R \) is then calculated using \( R = v \cos \theta \times T \). Plugging in our expression for \( T \), we arrive at the formula \( R = \frac{v^2 \sin 2\theta}{g} \). So option A is indeed the right choice! Projectiles have fascinated humans for centuries, from ancient siege weapons to modern space technology. Galileo Galilei made significant contributions to understanding projectile motion, establishing principles still used today. His experiments with inclined planes and balls led him to explore how angle and velocity affected distance, laying the groundwork for the study of dynamics and trajectories that continues to inspire scientists and engineers in today’s high-tech world!
