To find the horizontal distance traveled by the arrow, we can use the equations of projectile motion.
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Determine the initial velocity components: The initial velocity \( v_0 = 63.0 , \text{m/s} \) is at an angle of \( \theta = 45^\circ \).
The horizontal (x) and vertical (y) components of the initial velocity can be calculated using: \[ v_{0x} = v_0 \cdot \cos(\theta) \] \[ v_{0y} = v_0 \cdot \sin(\theta) \]
Given \(\theta = 45^\circ\): \[ v_{0x} = 63.0 \cdot \cos(45^\circ) = 63.0 \cdot \frac{\sqrt{2}}{2} \approx 44.6 , \text{m/s} \] \[ v_{0y} = 63.0 \cdot \sin(45^\circ) = 63.0 \cdot \frac{\sqrt{2}}{2} \approx 44.6 , \text{m/s} \]
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Determine the time of flight: The time of flight \( T \) can be found from the vertical motion. The total time of flight for a projectile launched and landing at the same height can be calculated using: \[ T = \frac{2 v_{0y}}{g} \] where \( g = 9.81 , \text{m/s}^2 \) (the acceleration due to gravity).
Plugging in the values: \[ T = \frac{2 \cdot 44.6}{9.81} \approx \frac{89.2}{9.81} \approx 9.09 , \text{s} \]
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Calculate the horizontal distance: The horizontal distance \( R \) (range) can be calculated using: \[ R = v_{0x} \cdot T \]
Substituting the values we found: \[ R \approx 44.6 , \text{m/s} \cdot 9.09 , \text{s} \approx 405.78 , \text{m} \]
Therefore, the horizontal distance traveled during the flight is approximately 405.78 meters.
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