Chapter 5 Quadratic Functions - 5.6 The Quadratic Formula And The Discriminant Worksheet Template Page 4
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GOAL
2
U
Q
F
R
L
SING THE
UADRATIC
ORMULA IN
EAL
IFE
2
In Lesson 5.3 you studied the model h = º16t
+ h
for the height of an object that
0
is dropped. For an object that is launched or thrown, an extra term v
t must be added
0
to the model to account for the object’s initial vertical velocity v
.
0
2
h = º16t
+ h
Models
Object is dropped.
0
2
h = º16t
+ v
t + h
Object is launched or thrown.
0
0
h = height
Labels
(feet)
t = time in motion
(seconds)
h
= initial height
(feet)
0
v
= initial vertical velocity
(feet per second)
0
The initial vertical velocity of a launched object can be positive, negative, or zero. If
the object is launched upward, its initial vertical velocity is positive (v
> 0). If the
0
object is launched downward, its initial vertical velocity is negative (v
< 0). If the
0
object is launched parallel to the ground, its initial vertical velocity is zero (v
= 0).
0
v
< 0
0
v
> 0
0
v
= 0
0
Solving a Vertical Motion Problem
E X A M P L E 5
Entertainment
A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet
above the ground and has an initial vertical velocity of 45 feet per second. The twirler
catches the baton when it falls back to a height of 5 feet. For how long is the baton in
the air?
S
OLUTION
2
Since the baton is thrown (not dropped), use the model h = º16t
+ v
t + h
with
0
0
v
= 45 and h
= 6. To determine how long the baton is in the air, find the value of t
0
0
for which h = 5.
2
h
v
h
= º16t
+
t +
Write height model.
0
0
2
5
= º16t
+ 45t +
6
h = 5, v
= 45, h
= 6
0
0
2
0 = º16t
+ 45t + 1
a = º16, b = 45, c = 1
º45 ± 2 0 8 9
t =
Quadratic formula
º32
t ≈ º0.022 or t ≈ 2.8
Use a calculator.
Reject the solution º0.022 since the baton’s time in the air cannot be negative.
The baton is in the air for about 2.8 seconds.
294
Chapter 5 Quadratic Functions
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