Projectile Motion archers question

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Two archers shoot arrows at different angles, 45.0 degrees and 60.0 degrees, both 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 calculated. The horizontal motion remains constant, while the vertical motion is analyzed using kinematic equations. The difference in landing distances reveals how angle affects projectile motion.
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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 verticle. The horizontal part would only have the equation v = d/t (since there is no acceleration in the horizontal component, and the verticle 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:

<br /> R = \frac{u^2 \sin{(2\theta)}}{g}<br />

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