Answer:
824 feet
Step-by-step explanation:
Given
[tex]h(t) = 16t^2[/tex]
[tex]H = 906[/tex] -- Initial Height
Required
Determine its height after 2 seconds
First, we calculate the distance covered in 2 seconds
[tex]h(t) = 16t^2[/tex]
[tex]h(2) = 16*2^2[/tex]
[tex]h(2) = 16*4[/tex]
[tex]h(2) = 64[/tex]
This means that the object moved a distance of 64feet
So, the object height is calculate by subtracting the distance covered from the object's initial height:
i.e.
[tex]Height = H - h(2)[/tex]
[tex]Height = 906- 64[/tex]
[tex]Height = 824[/tex]
Knowing the expression for the distance the object falls, the height of the object 2 seconds after it is dropped is 842 feet.
In first place, you know that the expression 16×t² represents the distance in feet the object falls after t seconds.
You want to know the height in feet of the object 2 seconds after it is dropped. For that, you must first calculate the distance traveled in 2 seconds. So you replace the time t by 2 seconds in the previous expression:
16×2²= 16×4= 64 feet
This means that the object moved a distance of 64 feet.
Knowing that the object is dropped from a height of 906 feet, the object height is calculate by subtracting the distance covered from the object's initial height:
Height= Initial Height - Distane covered
Height= 906 feet - 64 feet
Height= 842 feet
Finally, the height of the object 2 seconds after it is dropped is 842 feet.
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