CC Math 2
Quadratic Applications
1. A ball is thrown straight up with an initial velocity of 56 feet per second. The height of the ball t seconds after it is
thrown is given by the formula h(t) = 56t – 16t2.
a. What is the height of the ball after 1 second?
h(1) = 56(1) – 16(1)2 = 56 – 16 = 40 feet
b. What is the maximum height?
−b
−56
t = 2a = 2(−16) = 1.75
At 1.75 seconds, the maximum height is reached.
So h(1.75) = 49 feet
c. After how many seconds will it return to the ground?
When 56t – 16t2 = 0, the values for t represent when the height of the ball is 0.
Solving for t (by factoring out the GCF of 8t), t = 0 and 3.5.
So after 3.5 seconds, the ball will return to the ground.
2. An object is thrown upward into the air with an initial velocity of 128 feet per second. The equation
h(t) = 128t – 16t2 gives its height above the ground after t seconds. What is the height after 2 seconds? What is the
maximum height reached? For how many seconds will the object be in the air?
h(2) = 128(2) – 16(2)2 = 256 – 64 = 192 feet
The maximum height reached occurs at the vertex.
−b
−128
t = 2a = 2(−16) = 4
At 4 seconds, the maximum height is reached.
So h(4) = 128(4) – 16(4)2 = 512 – 256 = 256 feet
When 128t – 16t2 = 0, the values for t represent when the height of the object is 0.
Solving for t (by factoring out the GCF of 16t), t = 0 and 8.
So the object will be in the air for 8 seconds.
3. The length of a rectangle is three more than twice the width. Determine the dimensions that will give the rectangle a
total area of 27 m2.
L = 2W + 3
A = LW
(2W+3)(W) = 27
2W2 + 3W = 27
2W2 + 3W – 27 = 0
W = -4.5, W = 3
W = 3
L = 2(3) + 3 = 9
length = 9 feet; width = 3 feet
4. Lorenzo has 48 feet of fencing to make a rectangular dog pen. If a house were used for one side of the pen, what
would be the length and width for the maximum area?
L + 2W = 48
L = 48 – 2W
A = LW
A = (48 – 2W)(W)
A = 48W – 2W2
-2W2 + 48W = A
The vertex of this quadratic equation (W,A) represents the width that will yield the maximum area.
−48
W = 2(−2) = 12  L = 48 – 2(12) = 48 – 24 = 24
So the width is 12 feet, and the length is 24 feet.
5. What is the largest area that can be enclosed with 400 feet of fencing? What are the dimensions of the rectangle?
2L + 2W = 400
A = LW
2L = 400 – 2W
L = 200 – W
A = (200-W)(W)
-W2 + 200W = A
this quadratic equation represents the width (W) that yields a specific area (A)
If we find the vertex of this quadratic equation, we can find the width that will yield the maximum area.
−200
W=
= 100
2(−1)
So L = 200 – (100) = 100
width = 100 feet; length is 100 feet; area = 10,000 feet2
6. For the years of 1983 to 1990, the number of mountain bike owners m (in millions) in the US can be approximated
by the model m = 0.337t2 - 2.265t + 3.962, 3 ≤ t ≤ 10 where t = 3 represents 1983. In which year did 2.5 million
people own mountain bikes? In what year was the number of mountain bike owners at a minimum?
2.5 = 0.337t2 - 2.265t + 3.962
0 = 0.337t2 - 2.265t + 1.462
Solve for t.
a = .337
b = -2.265
c = 1.462
t=
−(−2.265)±√(−2.265)2 −4(.337)(1.462)
2(.337)
t=
2.265 ± √3.159449
.674
t=
2.265 ±1.777483896
.674
t = 5.998, t = .723
t ≈ 6  1986
t=
−(−2.265)
2(.337)
= 3.36 during the year 1983
7. A company’s weekly revenue in dollars is given by R(x) = 2000x − 2x2, where x is the number of items produced
during a week. What amount of items will produce the maximum revenue?
x=
−(2000)
2(−2)
= 500
500 items
8. The height in feet of a bottle rocket is given by h(t) = 160t −16t2 where t is the time in seconds. How long will it take
for the rocket to return to the ground? What is the height after 2 seconds?
160t – 16t2 = 0
16t(10 – t) = 0
t = 0, t = 10
10 seconds
h(2) = 160(2) – 16(2)2 = 320 – 64 = 256 feet
9. While playing catch with his grandson yesterday Tim throws a ball as hard as possible into the air. The height h in
feet of the ball is given by h = -16t2 + 64t + 8, where t is in seconds. How long will it take until the ball reaches the
grandson’s glove if he catches it at a height of 3 feet? What is the maximum height of the ball?
3 = -16t2 + 64t + 8
-16t2 + 64t + 5 = 0
t=
−(64)±√(64)2 −4(−16)(5)
2(−16)
−64±√4416
−32
t=
t = -.07, t = 4.07
It will take 4.07 seconds.
−b
−64
t = 2a = 2(−16) = 2
At 2 seconds, the ball is at its maximum height.
So h(2) = -16(2)2 + 64(2) + 8 = -64 + 128 + 8 = 72 feet