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?

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

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

[tex](5-4t)(5+4t) = 0[/tex]
[tex]5-4t = 0[/tex] and [tex]5+4t=0[/tex]
[tex]5=4t[/tex] and [tex]5=-4t[/tex]
[tex]t = \frac{5}{4} [/tex] and [tex]t = - \frac{5}{4} [/tex]

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

ernikhil ernikhil · Answer 2 · 2021-10-18T13:30:27+02:00

The acorn was moving through the air in the time interval of [tex]0\;\rm second[/tex] to [tex]1.25\;\rm seconds[/tex].

The given distance equation covered by acorn while falling from the tree is given as [tex]S(t)=-16t^2+25[/tex] which is time dependent.

Factorise the given equation by equating the given equation to zero,

[tex]-16t^2+25=0\\t^2=\dfrac{25}{16}\\t=\pm \dfrac{5}{4}\\t=1.25\;\rm seconds[/tex]

Hence, taking [tex]0\;\rm seconds[/tex] as the starting of fall, the acorn was moving through the air in the time interval of [tex]0\;\rm second[/tex] to [tex]1.25\;\rm seconds[/tex].

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