# Coffee Drop Experiment

## Coffee drop experiment

For the coffee filter the graph of position shows that it does not bounce out.

Although we have done a fit with a quadratic equation, the coefficient of the  is small. Thus the dependence of the position with time is almost linear.

$y(t)=At^2+Bt+C=0.1351+0.4247+0.0951$

thus $g=2*0.1351=0.27 (m/s^2)$ and $v_0=0.4247 (m/s)$

The gravitational acceleration that we found $g=0.27(m/s^2 )≪9.81$ (m/s^2 )=g_0$. This implies that the friction with air in this case is big and limits the velocity. From the fit of velocity we found a linear dependence that is almost horizontal$v(t)=0.08554 t+0.5594$where$v(0)=0.5594(m/s)$For the second drop of the coffee filter the data is almost the same$Y(t)=0.1351t^2+0.4247t+0.2951$thus$g=2*0.1351=0.27 (m/s^2)$and$v_0=0.4247 (m/s)$For the acceleration graph the fit is linear (the acceleration is not constant).$a=-0.4520t+0.5589$with an average value around$g=0.25 (m/s^2)\$

The explanation is simple: when we drop the coffee filter the friction with air is bigger than when we drop the ball. This friction force tends to limit the speed. Thus we find a smaller value for the experimental gravitational acceleration.