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