Questions: QUESTION 12 - 1 POINT An object is launched directly in the air at a speed of 144 feet per second from a platform located 12 feet in the air. The motion of the object can be modeled using the function f(t)=-16 t^2+144 t+12, where t is the time in seconds and f(t) is the height of the object. When, in seconds, will the object reach its maximum height? Do not include units in your answer.

QUESTION 12 - 1 POINT
An object is launched directly in the air at a speed of 144 feet per second from a platform located 12 feet in the air. The motion of the object can be modeled using the function f(t)=-16 t^2+144 t+12, where t is the time in seconds and f(t) is the height of the object. When, in seconds, will the object reach its maximum height? Do not include units in your answer.

Transcript text: QUESTION 12 - 1 POINT An object is launched directly in the air at a speed of 144 feet per second from a platform located 12 feet in the air. The motion of the object can be modeled using the function $f(t)=-16 t^{2}+144 t+12$, where $t$ is the time in seconds and $f(t)$ is the height of the object. When, in seconds, will the object reach its maximum height? Do not include units in your answer. Provide your answer below:

Solution

Solution Steps

Step 1: Identify the function and its components

The given function modeling the height of the object is: \[ f(t) = -16t^2 + 144t + 12 \]

Step 2: Determine the type of function

This is a quadratic function of the form $ f(t) = at^2 + bt + c $, where $ a = -16 $, $ b = 144 $, and $ c = 12 $.

Step 3: Find the time at which the maximum height occurs

For a quadratic function $ f(t) = at^2 + bt + c $, the maximum or minimum value occurs at $ t = -\frac{b}{2a} $. Since $ a < 0 $, the function has a maximum value.

Step 4: Calculate the time

Substitute $ a = -16 $ and $ b = 144 $ into the formula: \[ t = -\frac{144}{2(-16)} = -\frac{144}{-32} = 4.5 \]