# Ball Drop Experiment

## Ball drop experiment

Data was recorded for position and velocity as function of time. For position the floor is considered to be at a distance of 2 m. We have recoded data for 3 bounces of the ball.

Position of ball (meters) as a function of time (seconds). At t=0 seconds the ball is at maximum height.

The fit of the position of the ball after the first bounce was found to be quadratic

$Y(t)=4.762t^2-7.884t+4.115=\frac {gt^2}{2}+v_0 t+Y_0$

therefore the gravitational acceleration is

$g=2*4.762=9.524(m/s^2)$

and the initial speed after bounce

$v_0=-7.884(m/s)$ (upwards)

The numerical value of gravitational acceleration is credible especially because we have found the same value also from acceleration graph

Acceleration of ball (m/s^2) as function of time (seconds).

The fit of the acceleration is almost a horizontal line

$a=-0.4329t+9.841$

the slope $m=-0.4329 ≪ 1$ thus $a(0)=9.841m/s^2$ (initial acceleration)

For the second drop of the ball from the data we can find a little bigger numerical values for the gravitational acceleration and speed after bounce.

Position and velocity of the ball for the second drop. At time t=0 seconds the ball is at maximum height

For the position after the second drop the graph is also quadratic:

$Y(t)=\frac {gt^2}{2}+v_0 t+y_0=4.762t^2-7.884+4.115$

thus $g=9.524(m/s^2)$ and $v_0=-7.884 (m/s)$

The fit of the velocity is linear

$v=gt+v_0=9.518t-7.879 (m/s)$ thus $g=9.518 m/s^2 ≈9.524 m/s^2$ and $v_0=-7.879 m/s≈-7.884 (m/s)$