Answer
$5.4\times 10^{-13}\ m$
Work Step by Step
Given
The horizontal speed $v_{ox} = 3\times 10^6\ m/s$
The horizontal distance $x=1\ m$
Time require to travel in given distance is:
$t = \frac{x}{v_{ox}}$
$t = \frac{1\ m}{3\times 10^6\ m/s}$
$t = 3.33\times10^{-7}\ s$
The vertical displacement is given by:
$y =y_o+v_{oy}t + \frac{1}{2}gt^2$
$y =0+0(t )+ \frac{1}{2}gt^2$
$y = \frac{1}{2}gt^2$
$y = \frac{1}{2}\times 9.8\ m/s^2\times (3.33\times 10^{-7}\ s)^2$
$y=5.4\times 10^{-13}\ m$
