A student carried out an experiment to measure the acceleration due to gravity g - Leaving Cert Physics - Question 1 - 2021

The student measured the distance s using a measuring stick, ruler, or measuring tape. The distance was documented from the starting point to the bottom of the trapdoor where the object falls.

The student used a timer, starting it when the object was released from the top of the trapdoor and stopping it as soon as the object hit the ground. This ensures accurate measurement of the time taken for the fall.

To enhance the accuracy of the timing, the student adjusted the procedure to use a more precise timing device, such as an electronic timer, which reduces human reaction time errors. Additionally, placing the timer closer to the point of release can also help in measuring the fall more precisely.

Using the data where s = 0.4 m and t = 0.29 s:

1. Apply the formula:
   $g = \frac{2s}{t^2} = \frac{2 \times 0.4}{(0.29)^2}$
2. Calculate:
   $g = \frac{0.8}{0.0841} \approx 9.49 \, m/s^2$

Using the remaining data points:

• For s = 0.5 m, t = 0.32 s:
  $g = \frac{2 \times 0.5}{(0.32)^2} \approx 9.76 \, m/s^2$
• For s = 0.6 m, t = 0.38 s:
  $g = \frac{2 \times 0.6}{(0.38)^2} \approx 9.98 \, m/s^2$

Now average the three values calculated: $\text{Average} \, g = \frac{9.49 + 9.76 + 9.98}{3} \approx 9.74 \, m/s^2$

The student refrains from using a greater distance due to the potential for increased measurement errors. Longer distances may result in timing inaccuracies, as small delays in starting or stopping the timer can lead to significant discrepancies in the measured time, thereby affecting the calculated value for g.