Algebra 2
Chapter 5 Review - Word Problems
Name
1. A cliff diver in Acapulco, Mexico jumps from about 17 meters above the water. The function
y  4.9t 2  1.5t  17 models the height y in meters from the water of the cliff diver t seconds after he
jumps.
a. After how many seconds will the cliff diver hit the water? Round your answer to the nearest
tenth of a second.
 15
.  15
. 2  4(  4.9)(17)
t
2(  4.9)
 15
.  335.45
 9.8
t   171
. sec onds
t
t  2.02 sec onds
b. What is the height of the cliff diver 1 second after he jumps?
y   4.9(1) 2  15
. (1)  17
y  13.6 feet
2. The manufacturer of Knuckle Draggin' Snowboards found that when the unit price is x dollars, the
revenue R (in dollars) is: R  4 x 2  4000 x
a. What is the unit price needed to maximize the revenue?
x
 b  4000

 $500
2a
2(  4)
b. What is the maximum revenue?
R   4(500) 2  4000(500)
R   1,000,000  2,000,000
R  $1,000,000
3. In a softball game, an outfielder throws a ball to the shortstop. The path of the ball is modeled by the
equation y  0.002( x  195) 2  79.65 , where y is the height of the ball in feet after the ball has traveled
x feet horizontally. The shortstop catches the ball at the same height as the height at which the outfielder
released the ball.
a. What was the maximum height of the ball along its path? Round your answer to the nearest foot.
79.65  80 feet
b. How far was the shortstop from the outfielder at the time he caught the ball?
195(2)  390 feet
c. How high above the ground was the ball when it left the hand of the outfielder?
y   0.002(0  195) 2  79.65
y   76.05  79.65
y  3.6 feet