Answer
$x_f=1.23cm$
Work Step by Step
As we know that:
$x_f-x_{\circ}=v_{i}t+\frac{1}{2}at^2$
We know that: $a=g=9.8\frac{m}{s^2}$ , $x_{\circ}=0m$ ,$v_i=0\frac{m}{s}$, $t=50ms=50\times 10^{-3}s$
We plug in these known values to obtain:
$x_f-0=0(t)+\frac{1}{2}(9.8)(50\times 10^{-3})$
$x_f=4.9(50\times 10^{-3})$
$x_f=0.01225m$
$x_f=1.225cm$
$x_f=1.23cm$
