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Answer :
To find out how long the rocket was in the air before it returned to the ground, we use the given equation for height:
[tex]\[ h = 60t - 16t^2 \][/tex]
Here, [tex]\( t \)[/tex] is the time in seconds, and [tex]\( h \)[/tex] is the height in feet. The rocket will be on the ground when [tex]\( h = 0 \)[/tex]. So, we set the equation to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = 60t - 16t^2 \][/tex]
This can be rearranged as:
[tex]\[ 16t^2 - 60t = 0 \][/tex]
Factor out [tex]\( t \)[/tex]:
[tex]\[ t(16t - 60) = 0 \][/tex]
This equation gives us two potential solutions for [tex]\( t \)[/tex]. These solutions are:
1. [tex]\( t = 0 \)[/tex], which corresponds to the initial launch time.
2. [tex]\( 16t - 60 = 0 \)[/tex]
To solve the second equation:
[tex]\[ 16t - 60 = 0 \][/tex]
Add 60 to both sides:
[tex]\[ 16t = 60 \][/tex]
Divide by 16:
[tex]\[ t = \frac{60}{16} \][/tex]
Simplifying:
[tex]\[ t = 3.75 \][/tex]
Therefore, the rocket was in the air for a total of 3.75 seconds before it returned to the ground.
[tex]\[ h = 60t - 16t^2 \][/tex]
Here, [tex]\( t \)[/tex] is the time in seconds, and [tex]\( h \)[/tex] is the height in feet. The rocket will be on the ground when [tex]\( h = 0 \)[/tex]. So, we set the equation to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = 60t - 16t^2 \][/tex]
This can be rearranged as:
[tex]\[ 16t^2 - 60t = 0 \][/tex]
Factor out [tex]\( t \)[/tex]:
[tex]\[ t(16t - 60) = 0 \][/tex]
This equation gives us two potential solutions for [tex]\( t \)[/tex]. These solutions are:
1. [tex]\( t = 0 \)[/tex], which corresponds to the initial launch time.
2. [tex]\( 16t - 60 = 0 \)[/tex]
To solve the second equation:
[tex]\[ 16t - 60 = 0 \][/tex]
Add 60 to both sides:
[tex]\[ 16t = 60 \][/tex]
Divide by 16:
[tex]\[ t = \frac{60}{16} \][/tex]
Simplifying:
[tex]\[ t = 3.75 \][/tex]
Therefore, the rocket was in the air for a total of 3.75 seconds before it returned to the ground.
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