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Tag Archives: • 2) Hypothesis: In an If /Then statement highlight the purpose of the experiment.

November 28, 2025
November 28, 2025

Free-fall Lab

1) Introduction: Explain the theory behind this experiment in a paragraph between 150 and 250 words. (2 Points)

Suppose you are using external resources; include the reference. It would be best if you had any relevant formulas and explanations of each term. You may use the rich formula tools embedded here.

2) Hypothesis: In an If /Then statement, highlight the purpose of the experiment.

For instance: If two same shape objects with different masses are dropped from the same height, they will hit the ground simultaneously. (2 points)

Free-fall Lab

Post-lab section:

3) Attach your analysis here, including any table, chart, or plot image. (3 Points)

4) Attach the image of any table, chart, or plot here. (4 points)

Each part is 2 points.

Table 1 and the calculation of the percent error.

Table 2 and the calculation of the percent error.

5) Attach the image of samples of your calculation here. (2 points)

6) In a paragraph between 100 and 150 words, explain what you Learn. What conclusion can you draw from the results of this lab assignment? (2 points)

7) In one sentence, compare the results of the experiment with your Hypothesis. Why? (1 point)

8) Attach your response to the questions in the lab manual here. (4 points)

  • 1) Introduction: Explain the theory behind this experiment in a paragraph between 150 and 250 words. (2 Points),

  • 2) Hypothesis: In an If /Then statement highlight the purpose of the experiment.,

  • 3) Attach your analysis here including any table chart or plot image. (3 Points),

  • 6) In a paragraph between 100 and 150 words explain what you Learn. ,What conclusion can you draw from the results of this lab assignment? (2 points),

  • 7) In one sentence, compare the results of the experiment with your Hypothesis. Why? (1 point),


Comprehensive Answers (general)

1) Introduction (theory — ~180 words)

This experiment explores the kinematics of free-falling bodies and the determination of gravitational acceleration (g) using measured distances and times. For an object released from rest and falling under uniform gravity (neglecting air resistance), the displacement ss after time tt follows the kinematic equation

s=ut+12at2s = ut + \tfrac{1}{2} a t^2

where uu is initial velocity (here u=0u=0), and aa is acceleration (for free-fall, a=ga = g). Thus s=12gt2s = \tfrac{1}{2} g t^2. Rearranging gives g=2s/t2g = 2s/t^{2}. An alternative analysis plots ss versus t2t^{2}: the slope of that linear plot equals 12g\tfrac{1}{2}g, so g=2×(slope)g = 2\times(\text{slope}). Important experimental considerations: measurement uncertainty in distance and time (±∆s, ±∆t), reaction time when using a stopwatch, and air drag which becomes important for low-mass or high-surface-area objects. Percent error is used to compare measured gexpg_\text{exp} with the accepted value gtheog_\text{theo} (standard near Earth’s surface ~9.806 m/s²):

% error=∣gexp−gtheogtheo∣×100%.\%\text{ error} = \left|\frac{g_\text{exp}-g_\text{theo}}{g_\text{theo}}\right| \times 100\%.

References (example): classical kinematics textbooks or lab manual. Define each term when used (s = displacement in m, t = time in s, g = acceleration due to gravity in m/s²).


2) Hypothesis (If / Then)

If identical-shape objects of different masses are dropped from the same height in the same environment, then they will hit the ground at (nearly) the same time because gravitational acceleration is independent of mass when air resistance is negligible.


3) Analysis (tables, chart/plot, percent error — textual + worked example)

Below is a complete, general analysis you can copy into your report. Replace the sample data with your measured data and attach images of your actual tables/plots where requested.

How to analyze & what to plot

  • Collect several trials for each drop height (e.g., 0.5 m, 1.0 m, 1.5 m). For each trial record measured time tt.

  • Compute average time tˉ\bar{t} and t2t^2 for each height.

  • Table distance ss (m) vs average t2t^2 (s²). Fit a straight line; slope = ½ g.

Sample Table 1 — raw data and averages (text)

Height ss (m) Trial 1 t (s) Trial 2 t (s) Trial 3 t (s) Average tˉ\bar{t} (s) tˉ2\bar{t}^2 (s²)
0.50 0.32 0.33 0.31 0.320 0.1024
1.00 0.45 0.46 0.44 0.450 0.2025
1.50 0.55 0.56 0.54 0.550 0.3025

Linear fit method
Perform linear regression of ss (y-axis) vs t2t^2 (x-axis). For these sample numbers, slope ≈ 4.90 m/s² (this is 12g \tfrac{1}{2}g ), so measured gexp=2×slope≈9.80 m/s2g_\text{exp} = 2 \times \text{slope} \approx 9.80\ \text{m/s}^2.

Percent error calculation (Table 1 example)
Let gtheo=9.806 m/s2g_\text{theo}=9.806\ \text{m/s}^2, gexp=9.80 m/s2g_\text{exp}=9.80\ \text{m/s}^2

%error=∣9.80−9.8069.806∣×100%≈0.06%.\%\text{error} = \left|\frac{9.80-9.806}{9.806}\right|\times100\% \approx 0.06\%.

Table 2 — alternate method (g from each height)
Compute gg per height using g=2s/tˉ2g = 2s/\bar{t}^2.

Height ss (m) tˉ2\bar{t}^2 (s²) g=2s/tˉ2g = 2s/\bar{t}^2 (m/s²) % error
0.50 0.1024 2×0.5/0.1024=9.772\times0.5/0.1024 = 9.77 0.37%
1.00 0.2025 2×1.0/0.2025=9.882\times1.0/0.2025 = 9.88 0.77%
1.50 0.3025 2×1.5/0.3025=9.922\times1.5/0.3025 = 9.92 1.16%

Sample calculation (step-by-step)

  1. For height 1.00 m, avg time tˉ=0.450\bar{t}=0.450 s, tˉ2=0.2025\bar{t}^2 = 0.2025 s².

  2. g=2s/tˉ2=2(1.00)/0.2025=9.8765 m/s2.g = 2s/\bar{t}^2 = 2(1.00)/0.2025 = 9.8765\ \text{m/s}^2.

  3. Percent error = ∣(9.8765−9.806)/9.806∣×100%≈0.72%.|(9.8765-9.806)/9.806|\times100\% \approx 0.72\%.

Chart/plot instructions

  • Plot ss (vertical axis) vs t2t^2 (horizontal axis).

  • Add linear trendline and report slope ± uncertainty.

  • Compute g=2×slopeg = 2\times\text{slope} and include uncertainty propagation: if slope uncertainty is Δm\Delta m, then Δg=2Δm\Delta g = 2\Delta m.

Notes about attachments

  • I cannot attach images here; when you prepare the lab file: include (a) a scanned photo of Table 1, (b) the plotted graph (s vs t2t^2) with trendline and equation visible, (c) screenshots of calculation steps. Use high-contrast photos and crop tightly so numbers are legible.


4) (Requested attachment of image of table/chart) — guidance for your submission

  • Save your spreadsheet table as an image (PNG/JPG) or export the plotted chart as PNG.

  • Label axes (distance in m, time² in s²), include units and best-fit line equation (e.g., s=(4.90) t2+0.003s = (4.90)\,t^2 + 0.003).

  • Place these image files into your lab report PDF or upload where required.

November 4, 2025
November 4, 2025

Motion Lab Experiment

1) Introduction: Explain the theory behind this experiment in a paragraph between 150 and 250 words. (2 Points)

Suppose you are using external resources; include the reference. It would be best if you had any relevant formulas and explanations of each term. You may use the rich formula tools embedded here.

2) Hypothesis: In an If /Then statement, highlight the purpose of the experiment.

For instance: If two same shape objects with different masses are dropped from the same height, they will hit the ground simultaneously. (2 points)

Motion Lab Experiment

Post-lab section:

3) Attach your analysis here, including any table, chart, or plot image. (3 Points)

Motion Lab Experiment

4) Attach the image of any table, chart, or plot here. (4 points)

Each part is 2 points.

Table 1 and the calculation of the percent error.

Table 2 and the calculation of the percent error.

5) Attach the image of samples of your calculation here. (2 points)

 

6) In a paragraph between 100 and 150 words, explain what you Learn. What conclusion can you draw from the results of this lab assignment? (2 points)

 

7) In one sentence, compare the results of the experiment with your Hypothesis. Why? (1 point)

8) Attach your response to the questions in the lab manual here. (4 points)

  • 1) Introduction: Explain the theory behind this experiment in a paragraph between 150 and 250 words. (2 Points),
    2) Hypothesis: In an If /Then statement highlight the purpose of the experiment. ,
    3) Attach your analysis here including any table chart or plot image. (3 Points),
    4) Attach the image of any table chart or plot here. (4 points),
    5) Attach the image of samples of your calculation here. (2 points)


Answers (comprehensive, general)

1) Introduction (≈180 words)

This experiment investigates fundamental relationships of motion by measuring how a physical quantity (for example: acceleration due to gravity, period of a pendulum, or velocity during free fall) depends on controlled initial conditions. Classical kinematics and Newton’s second law form the theoretical basis. Kinematics relates displacement xx, initial velocity v0v_0, acceleration aa, and time tt with equations such as x=v0t+12at2x = v_0 t + \tfrac{1}{2} a t^2. If measuring free fall, the acceleration aa equals gravitational acceleration gg and the relevant formula for vertical displacement from rest is y=12gt2y = \tfrac{1}{2} g t^2. Newton’s second law, F=maF = ma, explains how net force produces acceleration and justifies controlled-force setups. When comparing measured (experimental) values to accepted (theoretical) values, compute percent error: %error=∣ experimental−theoretical ∣∣theoretical∣×100%\%\text{error} = \dfrac{|\,\text{experimental} – \text{theoretical}\,|}{|\text{theoretical}|}\times 100\%. Each variable should be defined: xx (displacement, m), v0v_0 (initial velocity, m/s), aa (acceleration, m/s²), tt (time, s), gg (9.81 m/s², standard). Sources for constants or additional theory (if used) should be cited (e.g., textbook or lab manual), and uncertainties from timing, measurement resolution, and systematic bias are discussed when interpreting results.


2) Hypothesis (If/Then)

If the experiment measures the motion of objects under controlled conditions (same release height and minimized air resistance), then the measured acceleration will be close to the theoretical value (e.g., g=9.81 m/s2g=9.81\ \text{m/s}^2) within experimental uncertainty, because classical kinematics predicts the time–distance relationship for uniformly accelerated motion.


3) Analysis (tables, charts, plots — general guidance & example)

Provide the raw data table, processed table (means, standard deviations), and at least one plot (e.g., displacement vs. t2t^2 for free fall, or velocity vs. time for constant acceleration). Below is a template and an example you can paste into a spreadsheet.

Table template (Raw Data):

Trial Height hh (m) Time tt (s) Notes
1
2
3
Mean

Processed table (example for free fall):

Height hh (m) Mean time tˉ\bar t (s) t2t^2 (s²) Computed gexp=2htˉ2g_\text{exp} = \dfrac{2h}{\bar t^2} (m/s²)
0.50 0.320 0.1024 9.77
1.00 0.451 0.2034 9.83

How to make useful plots:

  • For free-fall: plot hh (y-axis) vs t2t^2 (x-axis). A linear fit should have slope m=12gm = \tfrac{1}{2}g. Multiply slope by 2 to get gg.

  • For constant acceleration: plot velocity vs time; slope = acceleration.

Include error bars representing timing uncertainty (e.g., ± reaction time or instrument resolution) when possible.


4) Images of tables/charts/plots (explanation & submission notes)

I cannot attach files here, but follow these steps to create and attach images:

  1. Produce the table and plot in Excel/Google Sheets or Python/Matplotlib.

  2. Export the plot as PNG/JPG (File → Export/Download → PNG).

  3. For the report, include:

    • Table 1: Raw data and processed values.

    • Table 2: Summary (means, uncertainties).

    • Plot image: hh vs t2t^2 with linear fit and equation shown.

  4. Label each image as “Table 1”, “Table 2”, “Figure 1” in your submission.


5) Sample calculations (step-by-step example)

Provide a few worked-out samples; here are two typical calculations and the percent-error computation shown step-by-step.

Example: compute experimental gg from one measurement

  • Given: h=0.50 mh = 0.50\ \text{m}, mean measured time tˉ=0.320 s\bar t = 0.320\ \text{s}.

  • Formula: gexp=2htˉ2g_\text{exp} = \dfrac{2h}{\bar t^2}.

  • Compute tˉ2\bar t^2: 0.320×0.320=0.10240.320 \times 0.320 = 0.1024 (s²).

  • Compute numerator: 2h=2×0.50=1.002h = 2 \times 0.50 = 1.00 (m).

  • Divide: gexp=1.000.1024g_\text{exp} = \dfrac{1.00}{0.1024}.

    • Long division: 1.00÷0.1024≈9.7656251.00 \div 0.1024 \approx 9.765625 (m/s²).

  • Round appropriately: gexp≈9.77 m/s2g_\text{exp} \approx 9.77\ \text{m/s}^2.

Percent error calculation (step-by-step)

  • Theoretical gtheo=9.81 m/s2g_\text{theo} = 9.81\ \text{m/s}^2.

  • Experimental gexp=9.77 m/s2g_\text{exp} = 9.77\ \text{m/s}^2.

  • Absolute difference: ∣9.77−9.81∣=0.04|9.77 – 9.81| = 0.04.

  • Divide by theoretical: 0.04÷9.810.04 \div 9.81.