PHY2048 Labs — Lab Report 2: Free Fall & Acceleration Due to Gravity

Provided Data for Experiment 2 (Exp_2_data — Summer 2026)

Provided data for Exp 2 and instructions for data analysis and lab report.

Provided data for Exp 2 are given in Tables 1 & 2 (below).
Instruction for data analysis
1. Calculate t_(average) (s) and t²_(average) (s²) and record your calculated data in Tables 1 & 2 respectively.
2. Copy the data of t²_(average) (s²) and h (m) from Table 1 (for big ball), and paste them in the columns A and B in an Excel file.
3. Plot the data h vs t²_(average) and fit the data with a linear function y = mx + b where m is the slope, b is the intercept. Record the slope in Table 1. You can search on Google to find out how to do plot and fit with Excel.
4. Calculate g(exp.) and record it in Table 1.
5. Calculate % error between g(exp.) and g(accept.). Record the % error in Table 1.
6. Repeat steps (b) to (e) for small ball (the data are in Table 2). Record in Table 2 the fitting slope as well as all of your calculated data for small ball.
Instructions for lab report
1. Two figures of plotting and fitting for big and small balls must be included in your Exp 2 report.
2. Tables 1 & 2 with all the analyzed data must be included in your Exp 2 lab report.
3. Answers to the 4 questions at the end of Exp 2 lab manual must be included in your Exp 2 report.
4. The required other contents and format for your lab report can be found in the syllabus.

Table 1 — Free fall time of big ball (m₁) at heights h₁, h₂, h₃, h₄, h₅ (provided data)

h (m)	Trial 1 — t (s)	Trial 2 — t (s)	Trial 3 — t (s)	g(accept.)
1.655	0.580	0.579	0.572	9.8 m/s²
1.455	0.543	0.543	0.536
1.255	0.509	0.507	0.493
1.055	0.462	0.455	0.445
0.855	0.410	0.410	0.415

Table 2 — Free fall time of small ball (m₂) at heights h₁, h₂, h₃, h₄, h₅ (provided data)

h (m)	Trial 1 — t (s)	Trial 2 — t (s)	Trial 3 — t (s)	g(accept.)
1.655	0.581	0.580	0.581	9.8 m/s²
1.455	0.542	0.542	0.546
1.255	0.502	0.506	0.499
1.055	0.465	0.452	0.458
0.855	0.408	0.409	0.407

Experiment 2 — Completed Lab Report

🔒 Preview only — answers, data tables, plots, and calculations below are blurred.

Unlock the full completed lab report with every value, plot, calculation, and written answer.

Unlock full report

Pre-lab section

1. Introduction

Acceleration due to gravity is the acceleration experienced by a body when it falls freely under the influence of gravity alone, with no air resistance or other external forces acting on it. Near the surface of the Earth, this acceleration is denoted by g and has an approximate value of 9.81 m/s². Any object moving under this acceleration alone is said to be in free fall motion.

When a body is dropped from rest from a height h and the total time of fall t is measured, its motion can be analyzed using Newton's equation of motion: y = y₀ + v₀t + ½at² … (1), where y is the final position, y₀ is the initial position, v₀ is the initial velocity, and a is the constant acceleration.

In the case of free fall: y − y₀ = −h, v₀ = 0, and a = −g. Substituting these values into equation (1) gives h = ½gt² … (2). This equation relates the height of fall to the time taken and can be used to determine the value of g experimentally.

2. Hypothesis

Post-lab section

3. Analysis — h vs t²(average) plots

Plot 1 — Big ball (m₁): Linear fit y = mx, slope m = 4.99159848592069. The data points (t², h) are (0.170, 0.855), (0.206, 1.055), (0.253, 1.255), (0.293, 1.455), (0.333, 1.655). The line passes essentially through the origin, confirming h ∝ t².

Plot 2 — Small ball (m₂): Linear fit y = mx, slope m = 4.95901066792696. The data points (t², h) are (0.166, 0.855), (0.210, 1.055), (0.252, 1.255), (0.295, 1.455), (0.338, 1.655).

4. Table 1 — Big ball (m₁), completed

h (m)	Trial 1 (s)	Trial 2 (s)	Trial 3 (s)	t(average) (s)	t²(average) (s²)	Fitting slope	g(exp.)	g(accept.)	% error
1.655	0.580	0.579	0.572	0.577	0.333	4.99 m/s²	9.98 m/s²	9.8 m/s²	1.84%
1.455	0.543	0.543	0.536	0.541	0.293
1.255	0.509	0.507	0.493	0.503	0.253
1.055	0.462	0.455	0.445	0.454	0.206
0.855	0.410	0.410	0.415	0.412	0.170

Table 2 — Small ball (m₂), completed

h (m)	Trial 1 (s)	Trial 2 (s)	Trial 3 (s)	t(average) (s)	t²(average) (s²)	Fitting slope	g(exp.)	g(accept.)	% error
1.655	0.581	0.580	0.581	0.581	0.338	4.959 m/s²	9.918 m/s²	9.8 m/s²	1.20%
1.455	0.542	0.542	0.546	0.543	0.295
1.255	0.502	0.506	0.499	0.502	0.252
1.055	0.465	0.452	0.458	0.458	0.210
0.855	0.408	0.409	0.407	0.408	0.166

5. Sample calculations

For each height, t(average) = (Trial1 + Trial2 + Trial3) / 3, and t²(average) = [t(average)]². From h = ½gt², a plot of h vs t² is linear with slope m = g/2, so g(exp.) = 2m.

Big ball: m₁ slope = 4.99 → g(exp.) = 2(4.99) = 9.98 m/s². % error = |9.98 − 9.8| / 9.8 × 100 = 1.84%.

Small ball: m₂ slope = 4.959 → g(exp.) = 2(4.959) = 9.918 m/s². % error = |9.918 − 9.8| / 9.8 × 100 = 1.20%.

6. What I learned / Conclusion

From this laboratory experiment, I learned how the acceleration due to gravity can be determined experimentally by measuring the free-fall time of an object dropped from rest at various heights. By conducting multiple trials with both a large ball and a small ball, calculating the average fall times, and examining the relationship between height and time, I observed a linear trend that agrees with the theoretical equation h = ½gt².

The experimentally calculated values of gravitational acceleration for both balls were close to the accepted value of 9.8 m/s², with percentage errors of 1.84% and 1.20%, respectively. These relatively small errors indicate a high level of experimental accuracy. Furthermore, the results demonstrate that the mass of an object has a negligible effect on its motion in free fall when air resistance is minimal.

7. Comparison with the hypothesis

8. Answers to the lab manual questions

Q3. A body is dropped from rest from a height h = 6 m. Take g = 9.8 m/s².

(a) Time to reach the ground: t = √(2h/g) = √(2 × 6 / 9.8) = 1.11 s.

(b) Velocity just before impact: v = gt = 9.8 × 1.11 = 10.9 m/s.

Q4. Two balls of different mass dropped in a vacuum reach the bottom simultaneously because there is no air resistance. From t = √(2h/g), the time of fall depends only on h and g, not on mass, material, or density. Therefore both balls accelerate at the same rate g and arrive at the same instant.

Want the full unlocked Lab Report 2?

Get the complete completed report — every plot, calculation, and written answer — formatted and ready to submit, plus 1-on-1 help if you want to understand the work.

Unlock full report