Projectile Motion archers question

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SUMMARY

The discussion focuses on calculating the landing distances of two arrows shot by archers at different angles, specifically 45.0 degrees and 60.0 degrees, with the same initial speed. The arrow shot at 45.0 degrees lands 225 meters away. Using the formula for projectile range, R = (u^2 * sin(2θ)) / g, the ranges for both angles can be computed. The difference in landing distances between the two arrows is determined by calculating their respective ranges based on the initial speed and gravitational acceleration.

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  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions, specifically sine
  • Basic grasp of gravitational acceleration (g = 9.81 m/s²)
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  • Calculate the range of a projectile at 60.0 degrees using the formula R = (u^2 * sin(2θ)) / g
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Two archers shoot arrows in the same direction from the same place with the same initial speeds, but at different angles. One shoots at 45.0 degrees above the horizontal, while the other shoots at 60.0 degrees. If the arrow launched at 45.0 degrees lands 225m from the archer, how far apart are the two arrows when they land? assume that the arrows start at essentially ground level
 
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My old teacher taught us a trick for projectile motions. Divide it into two parts: horizontal and vertical. The horizontal part would only have the equation v = d/t (since there is no acceleration in the horizontal component, and the vertical component would have the the regular kinematic equations.

Remember that time is the same for both parts.
 
There is a very simple formula for calculating the range of a projectile in terms of the initial angle, starting speed and gravitational acceleration:

[tex] R = \frac{u^2 \sin{(2\theta)}}{g}[/tex]

compute the ranges of the two arrows and find out the difference in their ranges.
 

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