## Description

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 - 16t^{2}

Complete the table (attached) and discuss the interpretation of each point. Then answer the questions.

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## Explanation & Answer

Attached are the solutions in a word an pdf file. If you have any questions or need any corrections, please just let me know.

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 ...