Finding Acceleration Due to Gravity

Example: Find g from Data on a Falling Object

The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in introductory physics laboratory classes.

An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, the figure below. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

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. \(y_0 = 0;\) \(y = -1.0000\text{ m};\) \(t = 0.45173\text{ s};\) \(v_0 = 0.\)

2. Choose the equation that allows you to solve for \(a\) using the known values.

\(y = y_0 + v_0t + \cfrac{1}{2}at^2\)

3. Substitute 0 for \(v_0\) and rearrange the equation to solve for \(a\). Substituting 0 for \(v_0\) yields

\(y = y_0 + \cfrac{1}{2}at^2.\)

Solving for \(a\) gives

\(a = \cfrac{2(y \; -\; y_0)}{t_2}.\)

4. Substitute known values yields

\(a = \cfrac{2(-1.0000\text{ m} \; - \; 0)}{(0.45173\text{ s})^2} = -9.8010\text{ m/s}^2,\)

so, because \(a = -g\) with the directions we have chosen,

\(g = 9.8010\text{ m/s}^2.\)

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 \(9.80\text{ m/s}^2,\) so \(9.8010\text{ m/s}^2\) makes sense. Since the data going into the calculation are relatively precise, this value for \(g\) is more precise than the average value of \(9.80\text{ m/s}^2;\) it represents the local value for the acceleration due to gravity.

Check Your Understanding

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?

Solution

We know that initial position \(y_0 = 0,\) final position \(y = -30.0\text{ m}\), and \(a = -g = -9.80\text{ m/s}^2.\) We can then use the equation \(y = y_0 + v_0t + \frac{1}{2}at^2\) to solve for \(t\). Inserting \(a = -g\), we obtain

\(y = 0 + 0 \; - \; \frac{1}{2}gt^2\)

\(t^2 = \cfrac{2y}{-g}\)

\(t = ±\sqrt{\cfrac{2y}{-g}} = ±\sqrt{\cfrac{2(-30.0\text{ m})}{-9.80\text{ m/s}^2}} = ±\sqrt{6.12\text{ s}^2}\)

\(t = 2.47\text{ s} ≈ 2.5\text{ s}\)

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 Interactive Simulation: Graphing Equations

Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. \(y = bx\) to see how they add to generate the polynomial curve. See the video below by Megan Heine for a brief example on how to use the equation grapher which you can download via this link (367KB).