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) = -16t^2 + 25 ), where ( t ) is the number of seconds. for which interval of time is the acorn moving through the air?
\\( 0 < t leq \frac{5}{4} \\)
\\( 0 < t < \frac{5}{4} \\)
\\( t > \frac{5}{4} \\)
\\( -\frac{5}{4} < t < \frac{5}{4} \\)

Explanation:

Step1: Determine when the acorn hits the ground

The acorn is moving through the air until it hits the ground, which is when \( S(t) = 0 \). So we set up the equation:
\( 0=-16t^{2}+25 \)

Step2: Solve for \( t \)

First, rearrange the equation:
\( 16t^{2}=25 \)
Then divide both sides by 16:
\( t^{2}=\frac{25}{16} \)
Take the square root of both sides. Since time \( t \) cannot be negative in this context (we are talking about the time after the acorn starts falling), we have:
\( t = \sqrt{\frac{25}{16}}=\frac{5}{4} \)

Step3: Determine the interval for \( t \)

The acorn starts falling at \( t = 0 \) (when it begins to fall from the tree) and stops moving through the air when it hits the ground at \( t=\frac{5}{4} \). So the time interval when the acorn is moving through the air is \( 0 < t\leqslant\frac{5}{4} \) (we use \( \leqslant \) because at \( t = \frac{5}{4} \) it has just hit the ground, but the interval includes the time until it lands).

Answer:

\( 0 < t\leqslant\frac{5}{4} \) (corresponding to the first option: \( 0 < t\leqslant\frac{5}{4} \))