Question

Calculate the experimental acceleration due to gravity from the following data.
Time for travel 1: 0.795 s
Time for travel 2: 0.810 s
Time for travel 3: 0.782 s
Distance for fall: 300.0 cm

Answer 1

We use the formula:

$$g=\frac{2h}{t^2}$$

$$g(t_1)=2\frac{3}{0.795^2}=9.493\frac{m}{s^2}$$

$$g(t_2)=2\frac{3}{0.810^2}=9.145\frac{m}{s^2}$$

$$g(t_3)=2\frac{3}{0.782^2}=9.812\frac{m}{s^2}$$

The experimental acceleration due to gravity:

$$g=\frac{9.493+9.145+9.812}{3}=9.48\frac{m}{s^2}$$