If you launch a ball horizontally from a ramp at different heights, how will the horizontal range depend on the launch height? Sketch your predicted graph before reading on.
Syllabus 5.4 — Collect primary data to validate the relationships derived for projectile motion. Design, conduct, analyse and evaluate a first-hand investigation.
What we expect to observe based on the model
Practical Setup
Practical Setup Detailed
Aim: To determine the relationship between the launch height and the horizontal range of a projectile launched horizontally, and to compare the experimental results with the theoretical model.
Assumptions of the model:
For a horizontal launch from height $h$ with speed $v_x$:
If $v_x$ and $g$ are constant, then $R \propto \sqrt{h}$. A graph of $R$ versus $\sqrt{h}$ should be a straight line through the origin with gradient $v_x\sqrt{2/g}$.
Because $R \propto \sqrt{h}$, doubling the launch height does not double the range. Quadrupling the height doubles the range. This non-linear relationship is what the experiment will test.
A reproducible procedure for collecting valid data
Equipment:
Variables:
Procedure:
Ensure the landing area is clear. Use a ball that will not roll into traffic or off benches. Wear safety glasses if using a spring launcher.
Record your primary data here
| Trial | Height $h$ (m) | $\sqrt{h}$ (m½) | Range $R_1$ (m) | Range $R_2$ (m) | Range $R_3$ (m) | Mean Range $\bar{R}$ (m) | Uncertainty (m) |
|---|---|---|---|---|---|---|---|
| 1 | |||||||
| 2 | |||||||
| 3 | |||||||
| 4 | |||||||
| 5 |
Sample data (for comparison if you cannot perform the experiment):
| Height $h$ (m) | $\sqrt{h}$ (m½) | Mean Range $\bar{R}$ (m) |
|---|---|---|
| 0.10 | 0.316 | 0.32 |
| 0.20 | 0.447 | 0.45 |
| 0.30 | 0.548 | 0.55 |
| 0.40 | 0.632 | 0.64 |
| 0.50 | 0.707 | 0.71 |
These sample data assume $v_x \approx 1.0\ \text{m/s}$ and $g = 9.8\ \text{m/s}^2$.
Compare experiment to theory
Step 1 — Plot the graph
Plot $\bar{R}$ (vertical axis) versus $\sqrt{h}$ (horizontal axis). Draw a line of best fit.
Step 2 — Determine the gradient
The theoretical gradient is:
$m_{\text{theory}} = v_x \sqrt{\dfrac{2}{g}}$
From your graph, calculate the experimental gradient $m_{\text{exp}}$ using:
$m_{\text{exp}} = \dfrac{\Delta R}{\Delta \sqrt{h}}$
Step 3 — Compare
Calculate the percentage difference:
$\%\ \text{difference} = \dfrac{|m_{\text{exp}} - m_{\text{theory}}|}{m_{\text{theory}}} \times 100\%$
Step 4 — Calculate launch speed from data
Rearranging the gradient formula:
$v_x = m_{\text{exp}} \sqrt{\dfrac{g}{2}}$
Compare this calculated $v_x$ to any independent measurement of launch speed (e.g., from a motion sensor or timing gate).
A well-conducted investigation should yield a percentage difference under 10%. If your difference is larger than 15%, review your measurement technique and controlled variables.
Working Scientifically — identify and address limitations
Systematic errors:
Random errors:
Reliability:
Improvements:
A second validation using a different relationship
The theoretical time of flight is $t = \sqrt{2h/g}$. If you can measure $t$ independently (e.g., with a motion sensor or slow-motion video), you can validate:
$R = v_x \cdot t$
Rearranging: $v_x = R/t$. Calculate $v_x$ from each $(R, t)$ pair. If $v_x$ is approximately constant across all heights, this confirms that horizontal speed is unaffected by vertical motion — a fundamental assumption of the projectile model.
| Height $h$ (m) | Theoretical $t$ (s) | Measured $t$ (s) | $\%$ difference |
|---|---|---|---|
| 0.20 | 0.20 | ||
| 0.40 | 0.29 | ||
| 0.60 | 0.35 |
Theoretical times use $t = \sqrt{2h/g}$ with $g = 9.8\ \text{m/s}^2$.
Explain why the ball must be released from the same position on the ramp for every trial. What variable would be affected if the release position changed?
A student obtains a curved graph when plotting $R$ versus $h$, but a straight line when plotting $R$ versus $\sqrt{h}$. Explain why this observation validates the theoretical model $R = v_x\sqrt{2h/g}$. Include the physical meaning of the gradient in your answer.
Evaluate the assumption that air resistance is negligible in this experiment. Under what conditions would air resistance become significant? Describe how the experimental graph would deviate from the theoretical prediction if air resistance were significant, and explain the shape of this deviation.
The release position determines the gravitational potential energy converted to kinetic energy, which controls the horizontal launch speed $v_x$ (1 mark). If the release position changes, $v_x$ changes, making $v_x$ an uncontrolled variable (1 mark). Since $R \propto v_x$, any change in $v_x$ directly affects the range, confounding the relationship between $h$ and $R$ that we are trying to validate (1 mark).
The theoretical model predicts $R = v_x\sqrt{2/g} \cdot \sqrt{h}$, which is a linear relationship between $R$ and $\sqrt{h}$ passing through the origin (1 mark). The straight-line graph confirms this proportional relationship, while the curved $R$-$h$ graph reflects the square-root dependence (1 mark). The gradient of the $R$ versus $\sqrt{h}$ graph equals $v_x\sqrt{2/g}$ (1 mark), which is constant because $v_x$ and $g$ are constant in the experiment. This provides quantitative validation: the experimental gradient can be compared to the theoretical value (1 mark).
Air resistance is negligible for dense objects at low speeds, but becomes significant for light objects (e.g., ping-pong balls), large surface areas, or high speeds (1 mark). Air resistance opposes motion, reducing both horizontal speed and vertical acceleration below $g$ (1 mark). With air resistance, the experimental range would be less than the theoretical prediction at larger heights/speeds (1 mark). The graph of $R$ versus $\sqrt{h}$ would curve downward at higher values rather than remaining straight (1 mark), because air resistance increases with speed, causing greater fractional reduction in range as $h$ increases (1 mark).
Tick this box when you have finished all sections of this practical investigation.