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?

○ 0 < t ≤ 5/4

○ 0 < t < 5/4

○ t > 5/4

○ -5/4 < t < 5/4

Explanation:

Step1: Find when acorn hits ground

Set \( S(t) = 0 \) (ground level):
\( 0 = -16t^2 + 25 \)
\( 16t^2 = 25 \)
\( t^2 = \frac{25}{16} \)
\( t = \frac{5}{4} \) (time can’t be negative).

Step2: Determine time interval

At \( t = 0 \), acorn is at \( S(0) = 25 \) (starts falling). It moves until \( t = \frac{5}{4} \) (hits ground). Time \( t > 0 \) (since it starts falling at \( t = 0 \)) and \( t < \frac{5}{4} \) (before hitting ground). So interval is \( 0 < t < \frac{5}{4} \).

Answer:

\( 0 < t < \frac{5}{4} \) (the second option: \( 0 < t < \frac{5}{4} \))