How does the height of a drop affect the time it takes an object to hit the ground?
- Higher drops take longer.
- Physics gives us an equation to predict the relationship.
- Statistics lets us test the equation with data.
How does the height of a drop affect the time it takes an object to hit the ground?
Prediction: A plot of height versus time squared should be a straight line.
For an object released from rest with constant gravitational acceleration:
\[ h = \frac{1}{2}gt^2 \]
where:
Solving for time gives \(t = \sqrt{2h/g}\).
A simple linear regression has the form:
\[ y = \beta_0 + \beta_1 x + \varepsilon \]
For our falling-object data, set \(y=h\) and \(x=t^2\):
\[ h = \beta_0 + \beta_1t^2 + \varepsilon \qquad\text{with}\qquad \beta_1 \approx \frac{g}{2} \]
Thus, estimated gravity = 2 × regression slope. The intercept should be near zero.
This presentation uses 10 simulated drops, one from each height between 1 and 10 meters. Each falling time has a small simulated measurement error. These are not actual experimental measurements.
| Height (m) | Time (s) | Time squared (s²) |
|---|---|---|
| 1 | 0.454 | 0.207 |
| 2 | 0.604 | 0.365 |
| 3 | 0.786 | 0.617 |
| 4 | 0.908 | 0.825 |
| 5 | 1.008 | 1.015 |
| 6 | 1.114 | 1.242 |
| 7 | 1.195 | 1.427 |
| 8 | 1.259 | 1.586 |
| 9 | 1.353 | 1.832 |
| 10 | 1.436 | 2.061 |
Here is the R code that fits a straight line to height versus time squared:
# drops contains height, time, and time_squared model <- lm(height ~ time_squared, data = drops) summary(model) # Since slope = g/2, estimate gravitational acceleration: g_estimate <- 2 * coef(model)["time_squared"]
The lm() function estimates the slope and intercept.
The points lie close to the fitted straight line, as the free-fall equation predicts.
A residual is the difference between an observed height and the height predicted by the regression.
Residuals near zero mean the predictions are close to the simulated measurements.
Hover over a point to see the simulated measurement. You can also zoom and pan.
The small simulated timing errors make the estimates slightly different from the ideal values.