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An acrobat is fired from a cannon at 52 miles per hour at an angle of \( 55^{\circ} \) with respect to the horizontal. What are the horizontal and vertical components, \( v_{x} \) and \( v_{y} \), of the velocity vector \( \mathbf{v} \) ? \( \mathbf{v}=\left[v_{x}, v_{y}\right]=[ \) Ex:5.67,

Ask by Spencer Love. in the United States
Mar 18,2025

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The horizontal component \( v_{x} \) is approximately 29.8 mph, and the vertical component \( v_{y} \) is approximately 42.6 mph.

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Bonus Knowledge

To find the horizontal and vertical components of the velocity vector, you can use trigonometric functions. The horizontal component \( v_x \) can be calculated as \( v_x = v \cdot \cos(\theta) \) and the vertical component \( v_y \) as \( v_y = v \cdot \sin(\theta) \). Plugging in the values, where \( v = 52 \, \text{mph} \) and \( \theta = 55^\circ \): 1. \( v_x = 52 \cdot \cos(55^\circ) \approx 29.5 \, \text{mph} \) 2. \( v_y = 52 \cdot \sin(55^\circ) \approx 42.4 \, \text{mph} \) So, the velocity vector is approximately \( \mathbf{v} = [29.5, 42.4] \). When acrobats are launched from cannons, they follow a parabolic trajectory, similar to projectiles. This thrilling act combines physics and performance arts, providing not just entertainment but also a real-world application of kinematics and projectile motion that makes it fascinating for both the audience and the performers! Next time you're marveling at a cannon act, remember the physics behind it! With the right angle and velocity, these performers achieve breathtaking heights and distances. It’s a beautiful blend of artistry and mechanics—we're talking about Newton's laws of motion having a front-row seat at the circus!

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