This seems like pretty basic experiment, but I'm having a lot of trouble with it. Basically, I have two timer gates that measure time between two signals, and I drop metal ball between them. This way I'm getting distance traveled, and time. Ball is dropped from right above the first gate to make sure initial velocity is as small as possible (no way to make it 0 with this setup/timer). I'm assuming $v$ initial is $0 \frac{m}{s}$. Gates are $1$ meter apart.
Times are pretty consistent, and average result from dropping ball from $1.0$ meters is $0.4003$ seconds.
So now I have $3$ [constant acceleration] equations that I can use to get $g$.
$$d_{traveled} = v_{initial} * t + \frac{1}{2} * a * t^2$$$$d_{traveled} = v_{initial} . t + \frac{1}{2} a t^2$$ $$a = \frac{2*d}{t^2}$$$$a = \frac{2d}{t^2}$$ $$a = \frac{2*1.0}{.4003^2}$$$$a = \frac{2.1}{(.4003)^2}$$ $$a = 12.48 \frac{m}{s^2}$$
$$v_f^2 = v_i^2 + 2*a*d$$$$v_f^2 = v_i^2 + 2ad$$ $$a = \frac{v_f^2-v_i^2}{2d}$$ $$v_f = \frac{distance}{time}=\frac{1.0}{0.4003}=2.5 \frac{m}{s}$$ $$a = \frac{(2.5 \frac{m}{s})^2}{2*1.0 m}$$$$a = \frac{(2.5 \frac{m}{s})^2}{2.1 m}$$ $$a = 3.125 \frac{m}{s^2}$$
$$v_f = v_i + a*t$$$$v_f = v_i + at$$ $$a= \frac{v_f-v_i}{t}$$ $$a = \frac{2.5 m/s - 0}{ 0.4003 s}$$ $$a = 6.25 \frac{m}{s^2}$$
I'm getting three different results. And all of them are far from $9.8\frac{m}{s^2}$ . No idea what I'm doing wrong.
Also, if I would drop that ball from different heights, and plot distance-time graph, how can I get acceleration from that?
