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Sagot :
Answer:
Explanation:
To find the initial horizontal speed \( v_{0x} \) of the ball, we can use the following kinematic equations for projectile motion:
1. **Horizontal Motion**:
\[ x = v_{0x} \cdot t \]
where \( x \) is the horizontal distance traveled, \( v_{0x} \) is the initial horizontal speed, and \( t \) is the time of flight.
2. **Vertical Motion**:
\[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 \]
where \( y \) is the vertical displacement (which in this case would be the height of the building, but we are focused on the horizontal motion), \( v_{0y} \) is the initial vertical speed (which is \( v_0 \sin \theta \) for a projectile launched at an angle \( \theta \)), \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \ text per are also possible
To determine the initial horizontal speed \( v_{0x} \) of the ball, we use the following information and calculations:
Given:
- Angle of projection \( \theta = 37^\circ \)
- Horizontal distance \( x = 42.21 \) m
- Time of flight \( t = 2.34 \) seconds
First, find the initial vertical speed \( v_{0y} \):
\[ v_{0y} = v_0 \sin \theta \]
where \( v_0 \) is the initial speed of the ball.
Next, use the horizontal distance equation:
\[ x = v_{0x} \cdot t \]
Solve for \( v_{0x} \):
\[ v_{0x} = \frac{x}{t} \]
Now, let's find \( v_{0y} \):
\[ v_{0y} = v_0 \sin 37^\circ \]
To find \( v_0 \), we use the vertical motion equation to relate \( v_0 \) and the time of flight.
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