Kinematic Equations for Objects in Free Fall

Example 2.16 Find $\boldsymbol{g}$ 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 an introductory physics laboratory course.

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, Figure 2.43. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

Figure 2.43 Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.

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.

Figure 2.44

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. $y_{0}=0 ; y=-1.0000 \mathrm{~m} ; t=0.45173 ; v_{0}=0$.

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

$y=y_{0}+v_{0} t+\frac{1}{2} a t^{2}$

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

$y=y_{0}+\frac{1}{2} a t^{2}.$

Solving for $a$ gives

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

4. Substitute known values yields

$a=\frac{2(-1.0000 \mathrm{~m}-0)}{(0.45173 \mathrm{~s})^{2}}=-9.8010 \mathrm{~m} / \mathrm{s}^{2},$

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

$g=9.8010 \mathrm{~m} / \mathrm{s}^{2}.$

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