In a TV set, an electron beam moves with horizontal velocity of 4.2×10^7m/s across the cathode ray tube and strikes the screen, 45 cm away.The acceleration of gravity is 9.8 m/s2.How far does the electron beam fall while traversing this distance?Answer in units of m.
Find out how long it was travelling
Horizontal @ 4.2 x 10^7 m/s is constant
45 cm = .45 m
.45 m / 4.2 x 107 m/s = ........... seconds
Then to find the vertical distance
yf = yo + vo t - 1/2 (9.81 m/s2 ) t2
Yf is the distance you are looking for yo = 0 vo= 0
yf = -1/2 (9.81)(t2) using the 't' you found in the first step
