Question

1. analyzing the motion of a projectile a projectile is fired from a cliff 200 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 50 feet per second. the height h of the projectile above the water is modeled by h(x)=\frac{-32x^{2}}{(50)^{2}}+x + 200 where x is the horizontal distance of the projectile from the face of the cliff. (a) at what horizontal distance from the face of the cliff is the height of the projectile a maximum? (b) find the maximum height of the projectile. (c) at what horizontal distance from the face of the cliff will the projectile strike the water? (d) using a graphing utility, graph the function h, 0 ≤ x ≤ 200. (e) use a graphing utility to verify the solutions found in parts (b) and (c). (f) when the height of the projectile is 100 feet above the water, how far is it from the cliff?

Explanation:

Step1: Identify the quadratic - function form

The height function is $h(x)=\frac{-32x^{2}}{(50)^{2}}+x + 200=-\frac{32}{2500}x^{2}+x + 200$, which is in the form $y = ax^{2}+bx + c$ with $a=-\frac{32}{2500}$, $b = 1$, and $c = 200$.

Step2: Find the x - value of the vertex (for part a)

The x - coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substituting $a =-\frac{32}{2500}$ and $b = 1$ into the formula:
\[

$$\begin{align*} x&=-\frac{1}{2\times(-\frac{32}{2500})}\\ &=\frac{2500}{64}\\ & = 39.0625 \end{align*}$$

\]

Step3: Find the maximum height (for part b)

Substitute $x = 39.0625$ into the height function $h(x)$:
\[

$$\begin{align*} h(39.0625)&=-\frac{32}{2500}\times(39.0625)^{2}+39.0625 + 200\\ &=-\frac{32}{2500}\times1525.878906+39.0625 + 200\\ &=-19.53125+39.0625 + 200\\ &=219.53125 \end{align*}$$

\]

Step4: Find the x - value when $h(x)=0$ (for part c)

Set $h(x)=0$, so $-\frac{32}{2500}x^{2}+x + 200 = 0$. Multiply through by 2500 to get $-32x^{2}+2500x + 500000=0$. Divide by - 4 to simplify: $8x^{2}-625x - 125000 = 0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 8$, $b=-625$, and $c=-125000$:
\[

$$\begin{align*} x&=\frac{625\pm\sqrt{(-625)^{2}-4\times8\times(-125000)}}{2\times8}\\ &=\frac{625\pm\sqrt{390625 + 4000000}}{16}\\ &=\frac{625\pm\sqrt{4390625}}{16}\\ &=\frac{625\pm2095.38}{16} \end{align*}$$

\]
We take the positive root $x=\frac{625 + 2095.38}{16}\approx169.4$ (since distance cannot be negative).

Step5: For part f

Set $h(x)=100$, so $-\frac{32}{2500}x^{2}+x + 200=100$. Multiply through by 2500: $-32x^{2}+2500x+500000 = 250000$. Rearrange to $-32x^{2}+2500x + 250000=0$. Divide by - 4: $8x^{2}-625x - 62500=0$.
Using the quadratic formula $x=\frac{625\pm\sqrt{(-625)^{2}-4\times8\times(-62500)}}{16}=\frac{625\pm\sqrt{390625+2000000}}{16}=\frac{625\pm\sqrt{2390625}}{16}=\frac{625\pm1545}{16}$.
We get two solutions: $x_1=\frac{625 + 1545}{16}=\frac{2170}{16}=135.625$ and $x_2=\frac{625-1545}{16}=\frac{-920}{16}=-57.5$. We take the positive solution $x = 135.625$.

Answer:

(a) 39.0625 feet
(b) 219.53125 feet
(c) Approximately 169.4 feet
(f) 135.625 feet

(Note: Parts (d) and (e) require the use of a graphing - utility which is not shown in the step - by - step calculation but can be done using software like Desmos or a graphing calculator.)