Question

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

How far did the grapefruit drop?

Answer 1

Approximately
`2.0\; \rm m`, assuming that
`g = 9.81 \; {\rm m \cdot s^(-2)}` and that air resistance is negligible.

Step-by-step 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 &= (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 &= (1)/(2)\, a\cdot t^(2) + v_(0)\, t \\ &= (1)/(2) * 9.81\; \rm m\cdot s^(-2) * (0.64\; \rm s)^(2) + 0 \\ &\approx 2.0\; \rm m\end{aligned}`.