To determine if the ball will hit the wall, we analyze the motion of the ball in both horizontal and vertical directions. Here's the step-by-step solution:

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Step 1: Break the initial velocity into components

Given:

• Initial velocity $v_0 = 15.00 \, \text{m/s}$
• Angle $\theta = 30.00^\circ$

The horizontal and vertical components of velocity are: $v_{0x} = v_0 \cos\theta = 15.00 \cdot \cos(30^\circ) = 12.99 \, \text{m/s}$ $v_{0y} = v_0 \sin\theta = 15.00 \cdot \sin(30^\circ) = 7.50 \, \text{m/s}$

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Step 2: Time to reach the wall

The horizontal distance to the wall is $x = 11.00 \, \text{m}$. The horizontal motion is: $x = v_{0x} t$ Solving for time: $t = \frac{x}{v_{0x}} = \frac{11.00}{12.99} \approx 0.847 \, \text{s}$

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Step 3: Height of the ball when it reaches the wall

The vertical motion follows the equation: $y = y_0 + v_{0y}t - \frac{1}{2} g t^2$ Where:

• $y_0 = 2.00 \, \text{m}$ (initial height)
• $v_{0y} = 7.50 \, \text{m/s}$
• $g = 9.81 \, \text{m/s}^2$
• $t = 0.847 \, \text{s}$

Substituting values: $y = 2.00 + (7.50)(0.847) - \frac{1}{2}(9.81)(0.847)^2$ $y = 2.00 + 6.353 - 3.518 \approx 4.835 \, \text{m}$

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Step 4: Compare height to the wall

The height of the wall is $6.00 \, \text{m}$. The ball's height at $x = 11.00 \, \text{m}$ is $4.835 \, \text{m}$, which is less than the height of the wall.

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Conclusion

The ball will not hit the wall, as it passes underneath it.

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Let me know if you want me to expand on this! Here are 5 related questions:

1. How would the time of flight change if the angle of projection were increased?
2. What is the total time the ball stays in the air before hitting the ground?
3. How far horizontally does the ball travel before hitting the ground?
4. What is the maximum height the ball reaches during its trajectory?
5. How would the result change if air resistance was considered?

Tip: Always analyze both horizontal and vertical motions separately for projectile problems!