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The function
h(t)=-16t^(2)+144 represents the height,
h(t), in feet, of an object from the ground at
t seconds after it is dropped. A realistic domain (for time) for this function is:

-3 <= t <= 3

0 <= t <= 3

0 <= ℏ <= 144
all real numbers

The function $\newline$ $h(t) = -16t^{2} + 144$ represents the height, $\newline$ $h(t)$ , in feet, of an object from the ground at $\newline$ $t$ seconds after it is dropped. A realistic domain (for time) for this function is: $\newline$ $-3 \leq t \leq 3$ $\newline$ $0 \leq t \leq 3$ $\newline$ $0 \leq h \leq 144$ $\newline$ all real numbers

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Compare linear and exponential growth

Full solution

Q. The function

\newline

h(t) = -16t^{2} + 144

represents the height,

\newline

h(t)

, in feet, of an object from the ground at

\newline

t

seconds after it is dropped. A realistic domain (for time) for this function is:

\newline

-3 \leq t \leq 3

\newline

0 \leq t \leq 3

\newline

0 \leq h \leq 144

\newline

all real numbers

Understand function meaning: Understand the function and its physical meaning. $\newline$ The function $h(t) = -16t^2 + 144$ represents the height of an object in feet at time $t$ seconds after it is dropped. Since the object is dropped, it starts at a certain height and falls to the ground due to gravity. The domain of the function should reflect the time during which the object is in the air.
Analyze function coefficients: Analyze the coefficients of the function. $\newline$ The coefficient of $t^2$ is $-16$ , which reflects the acceleration due to gravity in feet per second squared (assuming the object is dropped on Earth). The constant term $144$ represents the initial height of the object in feet.
Determine realistic domain: Determine the realistic domain based on the physical context. $\newline$ Since time cannot be negative in this context, the domain cannot include negative values of $t$ . Therefore, the domain starts at $t = 0$ , when the object is dropped.
Find object landing time: Find when the object hits the ground. $\newline$ To find when the object hits the ground, we set $h(t)$ equal to $0$ and solve for $t$ : $\newline$ $0 = -16t^2 + 144$ $\newline$ $16t^2 = 144$ $\newline$ $t^2 = \frac{144}{16}$ $\newline$ $t^2 = 9$ $\newline$ $t = 3$ or $t = -3$ $\newline$ Since time cannot be negative, we only consider $t = 3$ . This is the time when the object hits the ground.
Establish domain for object: Establish the domain based on the time the object is in the air. The object is in the air from the time it is dropped $t = 0$ until it hits the ground $t = 3$ . Therefore, the realistic domain for the function is $0 \leq t \leq 3$ .

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