Q. 11 - Precalculus Enhanced with Graphing Utilities

Question

A projectile is fired from a cliff $200ft$ above the water at an inclination of $45°$ to the horizontal, with a muzzle velocity of $50ft/s$. The height h of the projectile above the water is modeled by

$h\left(x\right)=\frac{-32x^2}{{\left(50\right)}^2}+x+200$

where x is the horizontal distance of the projectile from the face of the cliff.

Part (a): At what horizontal distance from the face of the cliff is the height of the projectile a maximum?

Part (b): Find the maximum height of the projectile.

Part (c): At what horizontal distance from the face of the cliff will the projectile strike the water?

Part (d): Using a graphing utility, graph the function h,$0\le x\le 200$.

Part (e): Use a graphing utility to verify the solutions found in parts (b) and (c).

Part (f): When the height of the projectile is $100ft$ above the water, how far is it from the cliff?

Step-by-Step Solution

Answer

Part (a): The height of the projectile will be the maximum at a horizontal distance of $39ft$.

Part (b): The maximum height of the projectile is $219.5ft$.

Part (c): The projectile will strike the water at a horizontal distance of $170ft$from the face of the cliff.

Part (d): The required function is given below,

Part (e): When the height is $100ft$, the projectile is about $135.7ft$ from the cliff.

Part (f): When the height of the projectile is $100ft$, the projectile is at a distance of $135.7ft$from the cliff.

Part (a) Step 1. Given information.

Consider the given question,

Function for the height of the projectile is given below,

$h\left(x\right)=\frac{-32x^2}{{\left(50\right)}^2}+x+200$

Simplifying the given function,

$$\begin{gathered} h\left(x\right)=\frac{-32}{2500}{x}^2+x+200 \\ h\left(x\right)=\frac{-8}{625}{x}^2+x+200 ...... \left(i\right) \end{gathered}$$

Part (a) Step 2. Compare equation (i) with the standard quadratic equation.

Compare equation (i) with the standard quadratic equation,

$$\begin{gathered} a=-\frac{8}{625}, \\ b=1, \\ c=200 \end{gathered}$$

As $a<0$, the maximum value of h will be at the vertex. Then,

$$\begin{gathered} \frac{-b}{2a}=\frac{-1}{2\left({\displaystyle \frac{-8}{625}}\right)} \\ =\frac{625}{16} \\ =39ft \end{gathered}$$

Therefore, the height of the projectile will be the maximum at a horizontal distance of $39ft$.

Part (b) Step 1. Find the maximum height of the projectile.

Substitute $\frac{625}{16}$ for x in equation (i),

$$\begin{gathered} h\left(\frac{625}{16}\right)=\frac{-8}{625}{\left(\frac{625}{16}\right)}^2+\left(\frac{625}{16}\right)+200 \\ =-\frac{625}{32}+\frac{625}{16}+200 \\ =219.5 \end{gathered}$$

Therefore, the maximum height of the projectile is $219.5ft$.

Part (c) Step 1. Find the horizontal distance.

When the projectile strikes the water, the height will be $0$. Then,

$0=\frac{-8}{625}{x}^2+x+200 ...... \left(ii\right)$

Compare equation (ii) with the standard quadratic equation,

$$\begin{gathered} x=\frac{-1\pm \sqrt{11.24}}{2\left({\displaystyle \frac{-8}{625}}\right)} \\ x=170,-91.8 \end{gathered}$$

As the distance cannot be negative, then $x=-91.8ft$ cannot be negative.

Therefore, the projectile will strike the water at a horizontal distance of $170ft$ from the face of the cliff.

Part (d) Plot the function.

Plot the function for the interval $0\le x\le 200$,

Part (e): Step 1. Plot the function of part (b) and (c).

Consider the given question,

From the graph, the maximum height of the projectile is $219.5\mathrm{ft}$ and the distance where the projectile strike the water is $170ft$.

Part (f) Step 1. Find the distance from the cliff when h x = 100 .

Consider the given question, 

$h\left(x\right)=100$

Substitute the value in equation (i),

$$\begin{gathered} 100=\frac{-8x^2}{625}+x+200 \\ 0=\frac{-8x^2}{625}+x+100 ...... \left(iii\right) \end{gathered}$$

Compare equation (iii) with standard form of quadratic equation,

$$\begin{gathered} a=-\frac{8}{625}, \\ b=1, \\ c=100 \end{gathered}$$

Part (f) Step 2. Find the value of x.

Find the value of x, using the given values,

$$\begin{gathered} x=\frac{-1\pm \sqrt{6.12}}{2\left({\displaystyle \frac{-8}{625}}\right)} \\ x=135.7,-57.57 \end{gathered}$$

As the distance cannot be negative, then $x=-57.57$cannot be negative.

Therefore, when the height of the projectile is $100ft$, the projectile is at a distance of $135.7ft$ from the cliff.