An electron is ejected horizontally at a speed of 1.5 ✕ 106 m/s from the electron gun of a computer monitor. if the viewing screen is 35 cm away from the end of the gun, how far will the electron travel in the vertical direction before hitting the screen?

GivenHorizontal velocity, vxi = 1.5 *10^6 m/sDistance ,x =0.35 mFirst we will calculate the time takenx = vxi t0.35 m = (1.5 *10^6 m/s) tt =0.23 *10^-6 sThe vertical distance travelled by the electron isy =1/2 gt^2= 0.5 *9.8 m/s^2 *(0.23 *10^-6 s)^2= 0.25 *10^-12 m

Therefore, the answer is 0.25 *10^-12 m