Q. 31 - Precalculus Enhanced with Graphing Utilities

Question

Motion of a Golf Ball A golf ball is hit with an initial velocity of 130 feet per second at an inclination of 45° to the horizontal. In physics, it is established that the height h of the golf ball is given by the function 

$h\left(x\right)=\frac{-32x^2}{{130}^2}+x$

where x is the horizontal distance that the golf ball has traveled. 

(a) Determine the height of the golf ball after it has traveled 100 feet. 

(b) What is the height after it has traveled 300 feet? 

(c) What is h(500)? Interpret this value. 

(d) How far was the golf ball hit? 

(e) Use a graphing utility to graph the function h = h(x). 

(f) Use a graphing utility to determine the distance that the ball has traveled when the height of the ball is 90 feet. 

(g) Create a TABLE with TblStart = 0 and $∆Tbl$ = 25. To the nearest 25 feet, how far does the ball travel before it reaches a maximum height? What is the maximum height?

(h) Adjust the value of $∆Tbl$ until you determine the distance, to within 1 foot, that the ball travels before it reaches a maximum height.

Step-by-Step Solution

Answer

(a) 81.07 feet

(b) 129.59 feet

(c) 26.63 feet

(d) 528.13 feet

(e)

(f)

(g) 275 feet, 131.8 feet

(h) 264 feet

Part (a) Step 1. Given information

We have been given that A golf ball is hit with an initial velocity of 130 feet per second at an inclination of 45° to the horizontal. In physics, it is established that the height h of the golf ball is given by the function  

$h\left(x\right)=\frac{-32x^2}{{130}^2}+x$

where x is the horizontal distance that the golf ball has traveled.  

We have to Determine the height of the golf ball after it has traveled 100 feet.  

Part (a) Step 2. Substitute x = 100 in the given formula

$$\begin{gathered} h\left(100\right)=\frac{-32{\left(100\right)}^2}{{130}^2}+100 \\ = -\frac{320000}{16900}+100 \\ =-18.935+100 \\ \approx 81.07 \end{gathered}$$

Part (b) Step 1. Substitute x = 300 in the given formula

$$\begin{gathered} h\left(300\right)=\frac{-32{\left(300\right)}^2}{{130}^2}+300 \\ =-170.414+300 \\ \approx 129.59 \end{gathered}$$

Part (c) Step 1. Substitute x = 500 in the given formula

$$\begin{gathered} h\left(500\right)=\frac{-32{\left(500\right)}^2}{{130}^2}+500 \\ =-473.37+500 \\ \approx 26.63 \end{gathered}$$

This is the height of the golf ball after it has traveled 500 feet. 

Part (d) Step 1. Put h ( x ) = 0 in the given formula and solve for x

$$\begin{gathered} h\left(x\right)=\frac{-32x^2}{{130}^2}+x \\ 0=\frac{-32x^2}{{130}^2}+x \\ \frac{32x^2}{{130}^2}=x \\ \frac{32x}{{130}^2}=1 \\ x=\frac{{130}^2}{32} \\ x=528.13 \end{gathered}$$

Part (e) Step 1. Graph the given function

The function $h\left(x\right)=\frac{-32x^2}{{130}^2}+x$ is graphed as:

Part (f) Step 1. Draw a horizontal line from y = 90

The distance that the ball has traveled is 115.07 feet and 413.05 feet  when the height of the ball is 90 feet. 

Part (g) Step 1. Create a TABLE

x	h(x)
0	0
25	23.8
50	45.3
75	64.3
100	81.1
125	95.4
150	107.4
175	117
200	124.3
225	129.1
250	131.7
275	131.8
300	129.6
325	125
350	118
375	108.7
400	97
425	83
450	66.6
475	47.8
500	26.6
525	31
550	-22.8

The ball has traveled 275 feet before it reaches a maximum height of 131.8 feet.

Part (h) Step 1. Use formula h ( x ) = - 32 x 2 130 2 + x and create a table

x	h(x)
260	132
261	131.01
262	132.02
263	132.029
264	132.031
265	132.029
266	132.02
267	132.01
268	132

The ball has traveled 264 feet before it reaches a maximum height.