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 ≤ 5/4 0 < t < 5/4 t > 5/4 -5/4 < t < 5/4

Explanation:

Step1: Find when acorn hits ground

Set $S(t)=0$. So, $- 16t^{2}+25 = 0$.

Step2: Rearrange the equation

$16t^{2}=25$.

Step3: Solve for $t$

$t^{2}=\frac{25}{16}$, then $t=\pm\frac{5}{4}$.

Step4: Consider valid time

Since time $t\geq0$ in this context, we ignore $t =-\frac{5}{4}$. The acorn starts falling at $t = 0$ and hits the ground at $t=\frac{5}{4}$. So the time interval when it is moving through the air is $0 < t<\frac{5}{4}$.

Answer:

$0 < t<\frac{5}{4}$