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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) \( \frac{(v \sin \theta)^{2}}{\text { C) } \frac{v \sin 2 \theta}{2 g}} \begin{array}{l}\text { D) } \frac{(v \sin \theta)^{2}}{v g} \\ \text { E) } \frac{\nu}{2} \frac{\sin 2 \theta}{g}\end{array} \)

Ask by Robbins Huff. in the United States
Feb 04,2025

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The maximum horizontal displacement is \( \frac{v \sin 2\theta}{g} \).

Solución

The horizontal range \( R \) of a projectile launched from ground level with an initial velocity \( v \) at an angle \( \theta \) above the horizontal (ignoring air resistance) is given by the formula: \[ R = \frac{v^2 \sin 2\theta}{g} \] Where: - \( v \) is the initial velocity, - \( \theta \) is the launch angle, - \( g \) is the acceleration due to gravity. **Analyzing the Given Options:** Let's review the provided options and identify which one matches the standard formula for the horizontal range. 1. **Option A:** \[ \frac{v \sin 2\theta}{g} \] This option lacks a factor of \( v \) in the numerator. The correct formula has \( v^2 \), not just \( v \). 2. **Option B:** \[ \frac{(v \sin \theta)^2}{\text{...}} \] The option appears incomplete based on the formatting, but it doesn't directly match the standard formula. 3. **Option C:** \[ \frac{v \sin 2\theta}{2g} \] This is half of the correct range formula. 4. **Option D:** \[ \frac{(v \sin \theta)^2}{v g} = \frac{v (\sin \theta)^2}{g} \] This does not match the standard formula. 5. **Option E:** \[ \frac{v}{2} \cdot \frac{\sin 2\theta}{g} = \frac{v \sin 2\theta}{2g} \] Similar to Option C, this is also half of the correct formula. **Conclusion:** None of the options provided exactly match the standard formula for the horizontal range. However, **Option A** is the closest, likely intended to have the correct \( v^2 \) term but appears to be missing one factor of \( v \) in the numerator. Assuming a typographical error, **Option A** would be the correct choice. **Final Answer:** **Answer A) \( \dfrac{v \sin 2 \theta}{g} \)**

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The Deep Dive

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!

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