Aball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of time it takes to return to earth

In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

Parkway is the answer