Question

Quadratic Equations and Functions - Part 2

Answer 1

When the ball hits the ground, the height is 0. Substitute 0 for

h

.

h

=

−

16

t

2

−

10

t

+

200

0

=

−

16

t

2

−

10

t

+

200

−

16

t

2

−

10

t

+

200

=

0

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula,

x

=

−

b

±

√

b

2

−

4

a

c

2

a

. In this case, the variable is

t

rather than

x

.

a

=

−

16

,

b

=

−

10

, and

c

=

200

.

t

=

−

(

−

10

)

±

√

(

−

10

)

2

−

4

(

−

16

)

(

200

)

2

(

−

16

)

Simplify. Be very careful with the signs.

t

=

10

±

√

100

+

12800

−

32

=

10

±

√

12900

−

32

Use a calculator to find both roots.

t

is approximately

−

3.86

or

3.24

.

Consider the roots logically. One solution,

−

3.86

, cannot be the time because it is a negative number. The other solution,

3.24

seconds, must be when the ball hits the ground.

Answer

The ball hits the ground approximately

3.24

seconds after being thrown.

Answer 2

Answer:

See below

Step-by-step explanation:

Given function

• h(t) = 5 + 98t – 4.9t²

Part A

Filling in the table using the function, substitute t for each line:

time   |    elevation

0        |   5              

5        |   372.5      

10       |   495            

15       |   372.5      

20      |   5            

Part B

The graph is attached