# 2.7 Falling objects  (Page 5/9)

 Page 5 / 9

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?

Strategy

Draw a sketch.

We need to solve for acceleration $a$ . Note that in this case, displacement is downward and therefore negative, as is acceleration.

Solution

1. Identify the knowns. ${y}_{0}=0$ ; $y=–1\text{.0000 m}$ ; $t=0\text{.45173}$ ; ${v}_{0}=0$ .

2. Choose the equation that allows you to solve for $a$ using the known values.

$y={y}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}$

3. Substitute 0 for ${v}_{0}$ and rearrange the equation to solve for $a$ . Substituting 0 for ${v}_{0}$ yields

$y={y}_{0}+\frac{1}{2}{\text{at}}^{2}\text{.}$

Solving for $a$ gives

$a=\frac{2\left(y-{y}_{0}\right)}{{t}^{2}}\text{.}$

4. Substitute known values yields

$a=\frac{2\left(-1\text{.}\text{0000 m – 0}\right)}{\left(0\text{.}\text{45173 s}{\right)}^{2}}=-9\text{.}{\text{8010 m/s}}^{2},$

so, because $a=-g$ with the directions we have chosen,

$g=9\text{.}{\text{8010 m/s}}^{2}.$

Discussion

The negative value for $a$ indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of $9\text{.}{\text{80 m/s}}^{2}$ , so $9\text{.}{\text{8010 m/s}}^{2}$ makes sense. Since the data going into the calculation are relatively precise, this value for $g$ is more precise than the average value of $9\text{.}{\text{80 m/s}}^{2}$ ; it represents the local value for the acceleration due to gravity.

A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?

We know that initial position ${y}_{0}=0$ , final position $y=\text{−30}\text{.}\text{0 m}$ , and $a=-g=-9\text{.}{\text{80 m/s}}^{2}$ . We can then use the equation $y={y}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}$ to solve for $t$ . Inserting $a=-g$ , we obtain

$\begin{array}{lll}y& =& 0+0-\frac{1}{2}{\text{gt}}^{2}\\ {t}^{2}& =& \frac{2y}{-g}\\ t& =& ±\sqrt{\frac{2y}{-g}}=±\sqrt{\frac{2\left(-\text{30.0 m}\right)}{-9.80 m{\text{/s}}^{2}}}=±\sqrt{\text{6.12}\phantom{\rule{0.25em}{0ex}}{s}^{2}}=\text{2.47 s}\approx \text{2.5 s}\end{array}$

where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.

## Phet explorations: equation grapher

Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. $y=\text{bx}$ ) to see how they add to generate the polynomial curve.

## Section summary

• An object in free-fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration due to gravity $g$ , which averages
$g=9\text{.}{\text{80 m/s}}^{2}.$
• Whether the acceleration a should be taken as $+g$ or $-g$ is determined by your choice of coordinate system. If you choose the upward direction as positive, $a=-g=-9\text{.}\text{80 m}{\text{/s}}^{2}$ is negative. In the opposite case, $a=\mathrm{+g}=9\text{.}{\text{80 m/s}}^{2}$ is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate $+g$ or $-g$ substituted for $a$ .
• For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.

## Conceptual questions

What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down?

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