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?

Answer 1

$S(t) = -16t^2 + 25$ ⇒ Rewriting the equation gives
$S(t) = 25 - 16t^2$ ⇒ This is the 'difference of two squares'  form.

Equation S(t) to zero and Factorising S(t) we get

$(5-4t)(5+4t) = 0$
$5-4t = 0$ and $5+4t=0$
$5=4t$ and $5=-4t$
$t = \frac{5}{4}$ and $t = - \frac{5}{4}$

We know that the acorn falls from the height of 25 feet above the ground, it means the initial time when it falls is t = 0. The time when it lands on the ground is t = 1.25

So the acorn was in the air for 1.25 seconds

Answer 2

Answer:

Interval of time is 1.25 seconds.

Step-by-step explanation:

Given : 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.

To find :  For which interval of time is the acorn moving through the air?

Solution : The given equation is $S(t) = -16t^2 + 25$

where t is the time and S is the distance

Let the distance be zero i.e, at the ground.

$-16t^2 + 25=0$

$25-16t^2=0$

$(5)^2-(4t)^2=0$

$(5-4t)(5+4t)=0$

$5-4t=0,5+4t=0$

$t=\frac{5}{4},t=\frac{-5}{4}$

Time is  $t=\frac{5}{4}=1.25$

When the distance is 25 feet acron falls with initial time is zero

and when the distance is zero i.e, at the ground the time is 1.25 seconds

So, Interval of time is the acron moving through the air is 1.25-0=1.25 seconds.