91Ó°ÊÓ

In projectile motion, the horizontal range is the total distance a projectile travels along the horizontal axis before it hits the ground again. This is a crucial concept when solving problems related to objects launched into the air. To determine the horizontal range, we use the formula: \ \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] \ where \( R \) is the horizontal range, \( v_0 \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity. \ \
Given that the launch angle is 45°, the problem simplifies because \( \sin(90°) = 1 \). \ Thus, the formula simplifies to: \ \[ v_0 = \sqrt{ R \cdot g } \] \ For the problem, we were provided: \

• Range \( R = 107 \text{ m} \)
• Gravity \( g = 9.8 \text{ m/s}^2 \)

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Substituting these values gives us: \ \[ v_0 = \sqrt{ 107 \cdot 9.8 } \approx 32.4 \text{ m/s} \] \ This calculation gives us the initial velocity needed for the projectile to cover the given horizontal range.

The vertical position equation helps to find the height of the projectile at any given point in time. This is particularly useful when determining if the projectile clears an obstacle. The general formula is: \ \[ y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \] \ where: \

• \( y \) is the vertical position at time \( t \).
• \( y_0 \) is the initial vertical position.
• \( v_{0y} \) is the initial vertical velocity.
• \( g \) is the acceleration due to gravity.

\ Given: \

• Initial Height \( y_0 = 1.22 \text{ m} \)
• Initial Vertical Velocity \( v_{0y} = 32.4 \cos(45°) \approx 22.9 \text{ m/s} \)
• Time to reach fence \( t \approx 4.3 \text{ s} \)

\ Inserting these values into the vertical position equation, we get: \ \[ y = 1.22 + 22.9 \cdot 4.3 - \frac{1}{2} \cdot 9.8 \cdot (4.3)^2 \approx 8.96 \text{ m} \] \ This calculation indicates that at the horizontal distance of 97.5 meters, the vertical position of the ball is approximately 8.96 meters high. \ To determine if the ball clears the fence, compare this to the fence's height (7.32 m). Clearly, 8.96 m > 7.32 m, so the ball does clear the fence.

Initial velocity is key for solving projectile motion problems. It influences both the horizontal range and the trajectory of the projectile. As mentioned earlier, initial velocity \( v_0 \) was calculated using the formula: \ \[ v_0 = \sqrt{\frac{Rg}{\sin(2\theta)}} \] \ The steps involved in determining the initial velocity are: \

• Identify the horizontal range \( R \), launch angle \( \theta \), and gravity \( g \).
• Plug these values into the simplified formula for initial velocity at a 45° angle: \( v_0 = \sqrt{ R \cdot g } \).
• Compute the square root to find \( v_0 \). In this exercise, it was found to be approximately 32.4 m/s.

\ With this initial velocity, you can then move on to other calculations, such as determining the time to reach a certain horizontal distance and the vertical position of the projectile at that point. Each piece of the calculation helps build a complete picture of the projectile's flight path.