This is a problem of projectile motion where a shot passes just over a wall at a known height and horizontal distance. To solve this, we need to find the initial velocity ($v_0$) and the angle of projection ($\theta$).

Known Values

• Height of the wall ($h$) = 64 ft
• Horizontal distance to the wall ($x$) = 192 ft
• Acceleration due to gravity ($g$) = 32 ft/s² (standard value in feet).

Equations of Motion

1. Horizontal motion: $x = v_0 \cos\theta \cdot t$ $t = \frac{x}{v_0 \cos\theta}$

2. Vertical motion: $y = v_0 \sin\theta \cdot t - \frac{1}{2} g t^2$ At $x = 192 \, \text{ft}$, $y = 64 \, \text{ft}$.

Step 1: Substitute $t$ from Horizontal Motion into Vertical Motion

$y = v_0 \sin\theta \cdot \frac{x}{v_0 \cos\theta} - \frac{1}{2} g \left( \frac{x}{v_0 \cos\theta} \right)^2$ Simplify: $y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}$ Substitute $x = 192$, $y = 64$, and $g = 32$: $64 = 192 \tan\theta - \frac{32 \cdot 192^2}{2 v_0^2 \cos^2\theta}$

Step 2: Solve for $v_0^2$ and $\tan\theta$

Rewrite the equation: $64 = 192 \tan\theta - \frac{32 \cdot 192^2}{2 v_0^2 (1 + \tan^2\theta)}$ Let $k = \tan\theta$, so $1 + k^2 = \sec^2\theta$. Rearrange: $64 = 192 k - \frac{32 \cdot 192^2}{2 v_0^2 (1 + k^2)}$ $64 (1 + k^2) = 192 k (1 + k^2) - \frac{32 \cdot 192^2}{2 v_0^2}$ This equation involves both $v_0$ and $\theta$, so we solve numerically or iteratively for practical values.

Would you like me to proceed with the numerical solution or provide a symbolic simplification? Let me know!

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Expanding Understanding:

1. What is the effect of increasing the height of the wall on the angle of projection?
2. How does the horizontal distance influence the initial velocity?
3. Can we solve this problem if gravity is different (e.g., on the Moon)?
4. What happens if the shot must clear another obstacle beyond the wall?
5. How can air resistance affect the trajectory?

Tip: For projectile motion, always separate horizontal and vertical components to simplify equations.