Question

Please answer 30 points

Answer 1

Answer:

Initial height = 50ft

The flare will be at 58 ft and 1/2 seconds 1 seconds

it will hit the ground at 2.67 seconds

Step-by-step explanation:

h(t)= -16t^2 +24t+50

The initial height is 50

We want to find when h = 58

58= -16t^2 +24t+50

Subtract 58 from each side

58-58= -16t^2 +24t+50-58

0 =  -16t^2 +24t-8

Factor out -8

0 = -8(2t^2 -3t+1)

Factor inside the parentheses

0   = -8(2t -1)  (t-1)

Using the zero product property

2t-1 = 0    t-1 =0

t = 1/2        t=1

The flare will be at 58 ft and 1/2 seconds and 1 second

Both answers make sense,  one on the way up and one on the way down.

We need to find when the flare will hit the ground, h=0

0= -16t^2 +24t+50

Using the quadratic formula

-b ±sqrt(b^2-4ac)

--------------------------

2a

-24 ±sqrt(24^2-4(-16)50)

--------------------------

2(-16)

We get solutions for t

t≈-1.1703

t≈2.6703

Time cannot be zero, so it will hit the ground at 2.67 seconds

Answer 2

Answer:

Initial height = 50ft

The flare will be at 58 ft and 1/2 seconds and 1 second

it will hit the ground at 2.67 seconds

Step-by-step explanation:

h(t)= -16t^2 +24t+50

The initial height is 50

We want to find when h = 58

58= -16t^2 +24t+50

Subtract 58 from each side

58-58= -16t^2 +24t+50-58

0 =  -16t^2 +24t-8

Factor out -8

0 = -8(2t^2 -3t+1)

Factor inside the parentheses

0   = -8(2t -1)  (t-1)

Using the zero product property

2t-1 = 0    t-1 =0

t = 1/2        t=1

The flare will be at 58 ft and 1/2 seconds and 1 second

Both answers make sense,  one on the way up and one on the way down.

We need to find when the flare will hit the ground, h=0

0= -16t^2 +24t+50

Using the quadratic formula

-b ±sqrt(b^2-4ac)

--------------------------

2a

-24 ±sqrt(24^2-4(-16)50)

--------------------------

2(-16)

We get solutions for t

t≈-1.1703

t≈2.6703

Time cannot be zero, so it will hit the ground at 2.67 seconds