Question

The function f(t)=16t2 represents the distance (in feet) a dropped object falls in t seconds. The function g(t)=s0 represents the initial height (in feet) of the object. A necklace is dropped from a height of 25 feet. After how many seconds does the necklace hit the ground.

Answer 1

Answer: 1.25 seconds.

Explanation:

f(t) = (-16ft/s^2)*t^2 + g(t)

(i just wrote the units of the acceleration as ft per second squared, matching the gravitational acceleration in Earth)

this function will represent the height of a function that is dropped from an initial height g(t).

In this case, we have a necklace dropped from a height of 25 ft.

then g(t) = 25ft.

And the height equation will be:

f(t) = (-16ft/s^2)*t^2 + 25ft

The necklace will hit the ground hen f(t) = 0, then we can impose that and find the value of t.

f(t) = 0ft = (-16ft/s^2)*t^2 + 25ft

(16ft/s^2)*t^2 = 25ft

t = √(25ft/ (16ft/s^2)) = 1.25 s

Then the necklace will hit the ground after 1.25 seconds.