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?

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

Step-by-step explanation:

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

= -16t² + 25

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

-16t² + 25 = 0

16t² = 25

t² =

t =

t =   = 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.

f(x)=4

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