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NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level,

as a function of time is given by h ( t ) = − 4.9 t 2 + 346 t + 353 . For each of the answers below,

put your numerical answer, rounded to 1 decimal place, in the first box and the units, spelled out completely, in the second box.

User ZengJuchen
by
8.2k points

1 Answer

3 votes

Answer:

Explanation:

Certainly! Let's calculate the requested values based on the given function for the height of the rocket:

The height of the rocket at t = 2 seconds:

(

2

)

=

4.9

(

2

)

2

+

346

(

2

)

+

353

=

4.9

(

4

)

+

692

+

353

=

19.6

+

692

+

353

=

1025.4

h(2)=−4.9(2)

2

+346(2)+353=−4.9(4)+692+353=−19.6+692+353=1025.4

So, the height at t = 2 seconds is 1025.4 meters.

The height of the rocket at t = 5 seconds:

(

5

)

=

4.9

(

5

)

2

+

346

(

5

)

+

353

=

4.9

(

25

)

+

1730

+

353

=

122.5

+

1730

+

353

=

1960.5

h(5)=−4.9(5)

2

+346(5)+353=−4.9(25)+1730+353=−122.5+1730+353=1960.5

So, the height at t = 5 seconds is 1960.5 meters.

The height of the rocket at t = 10 seconds:

(

10

)

=

4.9

(

10

)

2

+

346

(

10

)

+

353

=

4.9

(

100

)

+

3460

+

353

=

490

+

3460

+

353

=

3323

h(10)=−4.9(10)

2

+346(10)+353=−4.9(100)+3460+353=−490+3460+353=3323

So, the height at t = 10 seconds is 3323 meters.

User Azr
by
8.0k points