To solve this problem, we need to find the horizontal distance $x$ at which the height of the projectile is maximum.

Given the equation for the height of the projectile: $h(x) = -\frac{32x^2}{50^2} + x + 190$ we recognize that this is a quadratic function of the form $h(x) = ax^2 + bx + c$, where:

• $a = -\frac{32}{50^2}$,
• $b = 1$,
• $c = 190$.

For a quadratic equation of the form $h(x) = ax^2 + bx + c$, the maximum (or minimum) value occurs at the vertex. The $x$-coordinate of the vertex is given by the formula: $x = \frac{-b}{2a}$

Step 1: Compute the coefficients

• $a = -\frac{32}{50^2} = -\frac{32}{2500}$
• $b = 1$

Step 2: Find the horizontal distance where the height is maximum

Substitute these values into the vertex formula: $x = \frac{-1}{2 \times \left(-\frac{32}{2500}\right)}$ Let me calculate this.The horizontal distance from the face of the cliff at which the height of the projectile is maximum is $x = 39.0625$ feet.

Would you like more details on how to interpret this result or ask any further questions?

Here are some related questions to consider:

1. What is the maximum height of the projectile?
2. How far horizontally does the projectile travel before hitting the water?
3. What is the time it takes for the projectile to reach its maximum height?
4. How does changing the muzzle velocity affect the maximum height and horizontal distance?
5. What would happen if the inclination angle was different from 45°?

Tip: The vertex of a quadratic function represents the point where the function reaches its maximum or minimum, depending on the sign of the leading coefficient.