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Quadratic functions can be used to describe the rate at which an object will fall and predict when it will land or how high it...Asked by L
Quadratic functions can be used to describe the rate at which an object will fall and predict when it will land or how high it may go. Create a quadratic function using the function y=-16t^2 to help predict when an object will return to the ground from being kicked, dropped, or thrown.
What does the 10 in my equation represent?
What does the 10 in my equation represent?
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GPT-4o mini
In the context of the quadratic function \( y = -16t^2 + h \), where \( h \) is the initial height from which the object is thrown (or kicked or dropped), the term \( 10 \) would represent the initial height \( h \) (in feet) from which the object is launched.
So, if you have the equation \( y = -16t^2 + 10 \), the \( 10 \) indicates that the object is initially 10 feet above the ground when it is kicked or thrown. The function describes how the height \( y \) of the object changes over time \( t \), where the term \( -16t^2 \) represents the effect of gravity (specifically, the acceleration due to gravity in feet per second squared).
To predict when the object will return to the ground, you would set \( y = 0 \) (when the object hits the ground) and solve for \( t \):
\[
0 = -16t^2 + 10
\]
Rearranging the equation gives:
\[
16t^2 = 10
\]
\[
t^2 = \frac{10}{16} = \frac{5}{8}
\]
\[
t = \sqrt{\frac{5}{8}} \approx 0.79 \text{ seconds}
\]
So, it would take approximately \( 0.79 \) seconds for the object to return to the ground if it was kicked from an initial height of 10 feet.
So, if you have the equation \( y = -16t^2 + 10 \), the \( 10 \) indicates that the object is initially 10 feet above the ground when it is kicked or thrown. The function describes how the height \( y \) of the object changes over time \( t \), where the term \( -16t^2 \) represents the effect of gravity (specifically, the acceleration due to gravity in feet per second squared).
To predict when the object will return to the ground, you would set \( y = 0 \) (when the object hits the ground) and solve for \( t \):
\[
0 = -16t^2 + 10
\]
Rearranging the equation gives:
\[
16t^2 = 10
\]
\[
t^2 = \frac{10}{16} = \frac{5}{8}
\]
\[
t = \sqrt{\frac{5}{8}} \approx 0.79 \text{ seconds}
\]
So, it would take approximately \( 0.79 \) seconds for the object to return to the ground if it was kicked from an initial height of 10 feet.
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