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You drop an object off the roof of a 441-foot building. The function h(t)=-16 t2+441 gives the object height h in feet above the ground after t seconds. After how many seconds does the object hit the ground?

You drop an object off the roof of a $441$ -foot building. The function $h(t)=-16 t 2+441$ gives the object height $h$ in feet above the ground after $t$ seconds. After how many seconds does the object hit the ground?

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. You drop an object off the roof of a

441

-foot building. The function

h(t)=-16 t 2+441

gives the object height

h

in feet above the ground after

t

seconds. After how many seconds does the object hit the ground?

Identify Equation: Identify the equation that gives the height of the object above the ground after $t$ seconds. $\newline$ The equation provided is $h(t) = -16t^2 + 441$ . $\newline$ We need to find the value of $t$ when the height $h(t)$ is $0$ , which represents the object hitting the ground.
Set Height Equal: Set the height equation equal to zero and solve for $t$ . $0 = -16t^2 + 441$ To solve for $t$ , we need to rearrange the equation and isolate $t$ .
Move $441$ : Move $441$ to the other side of the equation by subtracting it from both sides. $\newline$ $-16t^2 = -441$
Divide by $-16$ : Divide both sides of the equation by $-16$ to solve for $t^2$ . $\newline$ $t^2 = \frac{441}{16}$
Calculate $t^2$ : Calculate the value of $t^2$ . $t^2 = 27.5625$
Take Square Root: Take the square root of both sides to solve for $t$ . $t = \sqrt{27.5625}$
Calculate $t$ : Calculate the value of $t$ . $t \approx 5.25$ seconds $\newline$ This is the time it takes for the object to hit the ground.

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