# The maximum height attained by the ball.

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 3.1, Problem 64E

(a)

To determine

## To evaluate: The maximum height attained by the ball.

Expert Solution

The maximum height attained by the ball is 8.125ft .

### Explanation of Solution

Given:

A ball through across a playing field from a height of 5 feet above the ground at an angle of 45 to the horizontal at a speed of 20ft/s , the function of height (in feet) after x distance is given by y=32(20)2x2+x+5 .

Calculation:

The function of height (in feet) after x distance is given by,

y=32(20)2x2+x+5 (1)

The standard form of function,

f(x)=ax2+bx+c (2)

The maximum or minimum value of the function occurs at,

x=b2a (3)

If a>0 , then the minimum value is f(b2a) .

If a<0 , then the maximum value is f(b2a) .

From the equation (1) and (2),

a=32400=225b=1

Substitute 225 for a and 1 for b in the equation (3),

x=12×225=254=6.25

The function has maximum value,

a=225<0 .

The maximum value of the function at x=6.25 ,

f(6.25)=32400×(6.25)2+6.25+5=3.125+6.25+5=8.125

The maximum value of the function y=32(20)2x2+x+5 is 8.125 .

Thus, the maximum height attained by the ball is 8.125ft .

(b)

To determine

### To evaluate: The horizontal distance travelled by the ball when hit the ground.

Expert Solution

The horizontal distance travelled by the ball when hit the ground is 16.3ft .

### Explanation of Solution

Given:

A ball through across a playing field from a height of 5 feet above the ground at an angle of 45 to the horizontal at a speed of 20ft/s , the function of height (in feet) after x distance is given by y=32(20)2x2+x+5 .

Calculation:

The function of height (in feet) after x distance is given by,

y=32(20)2x2+x+5 (1)

The ball hits the ground is,

y=0 .

Substitute 0 for y in the equation (1),

32(20)2x2+x+5=0 (2)

The standard form of function,

ax2+bx+c=0 (3)

x=b±b24ac2a (4)

From the equation (2) and (3),

a=32400b=1c=5

Substitute 32400 for a , 1 for b and 5 for c in the equation (4),

x=1±124×(32400)×52×32400=1±1+20×(32400)2×32400=1±1+(3220)32200=1±1+(85)425

Simplify the above expression,

x=1±1350.16=1±2.60.16=1±1.610.16

For the value of x ,

x=11.610.16[Negativesign]=16.3,x=1+1.610.16[Positivesign]=3.8

Take positive value for the distance, the value of x is 16.3feet .

Thus, the horizontal distance travelled by the ball when hit the ground is 16.3ft .

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