The symbol 'g' can be used to refer to the acceleration due to gravity - Edexcel - GCSE Physics Combined Science - Question 4 - 2018 - Paper 1

Using the equation: $g = \frac{2h}{t^2}$ Substituting in the values:
h = 2.50 m and t = 0.74 s, we have: $g = \frac{2 \times 2.50}{(0.74)^2} = \frac{5.00}{0.5476} \approx 9.12 \, m/s^2$

To find the average of the three times: $t_{avg} = \frac{0.74 + 0.69 + 0.81}{3} = \frac{2.24}{3} \approx 0.747 \, s$ Rounded to two significant figures, the average value of time t is 0.75 s.

One way the students could improve their procedure is by using an electronic timer or a light gate to eliminate the reaction time involved in starting and stopping the stopwatch. This would lead to more precise measurements for the time taken by the ball to fall.

Using the equation for deceleration: $a = \frac{(v^2 - u^2)}{2s}$ where
v = 0 m/s (final speed), u = 15 m/s (initial speed), and s = 14 m (distance), we get: $a = \frac{(0^2 - (15)^2)}{2 \times 14} = \frac{-225}{28} \approx -8.04 \, m/s^2$ The deceleration of the car is approximately 8.04 m/s².