Answer:
Approximately [tex]2.0\; \rm m[/tex], assuming that [tex]g = 9.81 \; {\rm m \cdot s^{-2}}[/tex] and that air resistance is negligible.
Explanation:
If air resistance is negligible, the acceleration of this grapefruit during the fall would be constantly equal to [tex]g[/tex] (gravitational field strength.)
Let [tex]x[/tex] denote the distance that the grapefruit travelled during the fall. The following SUVAT equation gives an expression for [tex]x\![/tex] in terms of [tex]a[/tex], [tex]t[/tex], and [tex]v_{0}[/tex]:
[tex]\begin{aligned}x &= \frac{1}{2}\, a\cdot t^{2} + v_{0}\, t\end{aligned}[/tex].
In this question, the grapefruit was initially on a tree. Hence, assume that the initially velocity of this grapefruit was [tex]v_{0} = 0\; \rm m\cdot s^{-1}[/tex] right before the fall started.
If there was no drag on the grapefruit during the fall, the acceleration of this grapefruit would be equal to the gravitational field strength: [tex]a \approx 9.81\; \rm m\cdot s^{-2}[/tex].
The duration of the fall, [tex]t[/tex], has also been given. Hence, the distance that the grapefruit travelled during the fall would be:
[tex]\begin{aligned}x &= \frac{1}{2}\, a\cdot t^{2} + v_{0}\, t \\ &= \frac{1}{2} \times 9.81\; \rm m\cdot s^{-2} \times (0.64\; \rm s)^{2} + 0 \\ &\approx 2.0\; \rm m\end{aligned}[/tex].
