Physics help please?

Question

In a TV set, an electron beam moves with horizontal velocity of 3.4x10^7 m/s across the cathode ray tube and strikes the screen, 45 cm away. The acceleration of gravity is 9.8 m/s^2 .
How far does the electron beam fall while traversing this distance? Answer in units of m.

Any help appreciated thanks.

Answer 1

s = s0 +v0t+(1/2)at^2
horizontally, s0 = =, a = 0, so
45 cm = (3.4*10^7 m/sec)*t = 0.45 m, so
t = 0.45/(3.4*10^7) sec
t = (4.5*10^-8)/3.4 sec

vertically, s0 = 0, v0 = 0, so
s = (1/2)(9.8 m/sec^2)((4.5*10^-8)/3.4 sec)^2