One exercise says to drop something from a certain height and measure the time it takes for it to hit the floor and then calculate the gravitational acceleration, but can you show me how to calculate with the uncertainties I got?
The times i got was 0.20, 0,18s, 0,13s.. ten times. So $\overline{t}=\frac{0.20 + 0.18..}{10}$ and $\Delta t = \frac{max-min}{2}$ so the uncertainty in time is $(0,203 \pm0,09)s$ and the height is $(0,715\pm 0,05)m$. (Im not sure about the number of significant digits?)
To calculate the gravitational acceleration we have $s=V_0t + \frac{1}{2}at^2 \Rightarrow a =\frac{2s}{t^2}$. So is
$$a=\frac{2\cdot (0,715\pm0,05)m}{(0,203\pm0,09)s^2}$$
correct? And how do I calculate that? I appreciate your time and help.
You measured $n=10$ times, lets call them $t_i$ with $ i=1,\dots,10$. The mean time is given by $$ \bar{t}:= \frac{1}{n} \sum_{i=1}^n t_i = \frac{0.20+0.19+0.13+\dots}{10}s \,. $$ The uncertainty on the mean time is given by $$ \sigma_{\bar{t}} = \sqrt{\frac{1}{n(n-1)} \sum_{i=1}^n (\bar{t}-t_i)^2} \,. $$
The height was measred to be $h=0.715m$ and its uncertainty $\sigma_h=0.05m$.
Now we have to compute gravitational accerleration $$ a=\frac{2h}{\bar{t}^2} = \frac{2\cdot 0.715m}{\bar{t}^2} \,. $$ Its uncertainty can be calculated via error propagation $$ a=a(h,t) \implies \sigma_a = \sqrt{\left(\frac{\partial a}{\partial h} \sigma_h\right)^2 +\left(\frac{\partial a}{\partial t} \sigma_{\bar{t}}\right)^2} = \sqrt{\left(\frac{2}{\bar{t}^2} \sigma_h\right)^2 +\left(\frac{4h}{\bar{t}^3} \sigma_{\bar{t}}\right)^2} \,. $$