Answer:
[tex]0<t<\frac{5}{4}[/tex]
Step-by-step explanation:
We are given that,
The fall of the acorn is represented by the function,
[tex]S(t) =-16t^2+25[/tex]
Equating S(t) to 0 gives,
[tex]0=-16t^2+25[/tex]
i.e. [tex]16^2=25[/tex]
i.e. [tex]t^{2}=\frac{25}{16}[/tex]
i.e. [tex]t=\pm \frac{5}{4}[/tex]
Since, time cannot be negative.
So, [tex]t=\frac{5}{4}[/tex] i.e. 1.25 seconds is the time when the acorn will reach the ground.
Thus, from the figure below, we see that,
The interval in which the acorn is in the moving air is [tex]0<t<\frac{5}{4}[/tex].
