Answer
$d_y=1.5mm$
Work Step by Step
We know that,
$F_{net}=ma_y$
Here,
$F_{net}=F-F_g=4.5\times10^{-16}-(9.11\times10^{-31}\times9.8)=4.49\times10^{-16} N$
Thus, $ma_y=4.49\times10^{-16} $
This leads to:
$a_y=\frac{4.49\times10^{-16}}{9.11\times10^{-31}}=4.94\times10^{14}\frac{m}{s^2}$
As $d_x=v_xt$
or $t=\frac{d_x}{v_x}=\frac{0.03m}{1.2\times10^7\frac{m}{s}}=2.5\times10^{-9}s$
Also, for vertical distance
$d_y=vi_yt+0.5a_yt^2$
As ${v_iy}=0$ (that is initial vertical velocity is zero), the above equation becomes:
$d_y=0.5 a_y t^2$
Putting the values,
$d_y=0.5 (4.94\times10^{14})(2.5\times10^{-9})^2$
$d_y=0.0015m=1.5mm$
