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?

A.0
B.0
C.T>5/4

D.-5/4

Answer 1

so hmm check the picture below... that'd be the graph for an initial velocity scenario

so... in short, when the acorn hits the ground, the value for "y" is 0, so, what is "t" value at that point? well, let's do so, set y = 0

$\bf s(t)=-16t^2+25\implies 0=-16t^2+25\implies -25=-16t^2 \\\\\\ \cfrac{-25}{-16}=t^2\implies \sqrt{\cfrac{25}{16}}=t\implies \cfrac{\sqrt{25}}{\sqrt{16}}=t\implies \cfrac{5}{4}=t\implies 1\frac{1}{4}=t$

so.. it'd  be  $\bf 1\frac{1}{4}\ seconds$  or 1.25 seconds

Answer 2

Answer:

$[0,\frac{5}{4})$ or $0\leq t<\frac{5}{4}$

Step-by-step explanation:

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 number of seconds.

We need to find the time for which acorn moving through the air.

If acorn in air then S(t)>0

$-16t^2+25>0$

$-16t^2>-25$

$t^2<\dfrac{25}{16}$

Taking square root both sides

$t<\dfrac{5}{4}$

Interval of t, $[0,\dfrac{5}{4})$

Hence, t is less than 5/4 and greater than 0.