Question

An acorn falls from the branch of a tree to the ground 25 feet below. The distance, S, that the acorn is from the ground as it falls is represented by the equation S(t) = –16t2 + 25, where t is the number of seconds. For which interval of time is the acorn moving through the air?

Answer 1

The equation  is the initial distance of the acorn at time t. So in order to find the time for the acorn to reach the ground, we will just equate the equation above to zero  and find t. We are going to get the positive value since there is no such thing as negative time. So for the acorn to reach the ground, it would take 1.25 seconds.

Answer 2

Acorn will remain in the air between the interval of 0 to 1.25 seconds.

Explanation:

The given expressions that represents the height s, that the acorn is from the ground at time t

`S_((t))` = -16t² + 25

Now we have to calculate the time t for which the distance S become Zero.

-16t² + 25 = 0

16t² = 25

t² =
`(25)/(16)`

t =
`\sqrt{(25)/(16) }`

t =
`(5)/(4)` = 1.25 seconds

Therefore, acorn will be in air for 1.25 seconds before touching the ground.

In other words acorn will remain in air between the interval of 0 to 1.25 seconds.