an acorn falls from the branch of a tree to the ground 25 feet below. the distance, s, that the acorn is…

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?\n0 < t ≤ 5/4\n0 < t < 5/4\nt > 5/4\n-5/4 < t < 5/4

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?\n0 < t ≤ 5/4\n0 < t < 5/4\nt > 5/4\n-5/4 < t < 5/4

Answer

Explanation:

Step1: Find when acorn hits ground

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

Step2: Solve the equation

Rearrange to get $16t^{2}=25$, then $t^{2}=\frac{25}{16}$. Taking square - roots, $t=\pm\frac{5}{4}$. Since time $t\geq0$ in this context, we discard $t =-\frac{5}{4}$.

Step3: Determine time interval in air

The acorn starts falling at $t = 0$ and hits the ground at $t=\frac{5}{4}$. So the time it is in the air is $0\lt t\lt\frac{5}{4}$.

Answer:

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