High School

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Mr. Alim's science class launched a toy rocket from ground level with an initial upward velocity of 60 feet per second. The height [tex]h[/tex] of the rocket in feet above the ground after [tex]t[/tex] seconds is modeled by the equation [tex]h = 60t - 16t^2[/tex]. How long was the rocket in the air before it returned to the ground?

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.

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