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Using a Quadratic Equation Problems
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per
second. The distance s (in feet) of the ball from the ground after t seconds is
s (t) = 112 + 96t - 16t2
Complete the table and discuss the interpretation of each point.
t
0

s(t)
2

s(0)=112+96∗0−16∗0
s (0)=112

Interpretation
At t = 0, the ball is just about to be
thrown, its height is 112 ft. This is
also the height of the building, which
makes sense because that is where
the ball is thrown from.
At t = 0.5, the ball has been in the air
for 0.5 seconds and it now at a height
of 156 ft and traveling upward.

0.5

s(0.5)=112+96∗0.5−16∗0.52
s(0.5)=112+48−4
s(0.5)=156

1

s(1)=112+96∗1−16∗12
s (1)=112+96−16
s(1)=192

At t = 1, the ball has been in the air
for 1 seconds and it now at a height
of 192 ft.

2

s( 2)=112+96∗2−16∗22
s( 2)=112+192−64
s (2)=240

At t = 2, the ball has been in the air
for 2 seconds and it now at a height
of 240 ft

100=112+96t−16t 2
0=12+96t−16t 2
−96±√ 96 2−4∗−16∗12
t=
2∗−16
−96±√ 9984
t=
−32
t =−0.122 and 6.122

100

The ball will reach a height of ...