Question

A grapefruit falls from a tree and hits the ground 0.64 s later.

Answer 1

Answer:

Approximately $2.0\; \rm m$, assuming that $g = 9.81 \; {\rm m \cdot s^{-2}}$ 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 $g$ (gravitational field strength.)

• Let $a$ denote the acceleration of this grapefruit.
• Let $t$ denote the duration of the fall.
• Let $v_{0}$ denote the initial velocity of this grapefruit (right before the fall started.)

Let $x$ denote the distance that the grapefruit travelled during the fall. The following SUVAT equation gives an expression for $x\!$ in terms of $a$, $t$, and $v_{0}$:

$\begin{aligned}x &= \frac{1}{2}\, a\cdot t^{2} + v_{0}\, t\end{aligned}$.

In this question, the grapefruit was initially on a tree. Hence, assume that the initially velocity of this grapefruit was $v_{0} = 0\; \rm m\cdot s^{-1}$ 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: $a \approx 9.81\; \rm m\cdot s^{-2}$.

The duration of the fall, $t$, has also been given. Hence, the distance that the grapefruit travelled during the fall would be:

$\begin{aligned}x &= \frac{1}{2}\, a\cdot t^{2} + v_{0}\, t \\ &= \frac{1}{2} \times 9.81\; \rm m\cdot s^{-2} \times (0.64\; \rm s)^{2} + 0 \\ &\approx 2.0\; \rm m\end{aligned}$.