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.

Guest Oct 5, 2020

#1**0 **

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 10^{7 }m/s = ........... seconds

Then to find the vertical distance

yf = yo + vo t - 1/2 (9.81 m/s^{2} ) t^{2 }

^{Yf is the distance you are looking for yo = 0 vo= 0 }

yf = -1/2 (9.81)(t^{2}) using the 't' you found in the first step

Guest Oct 5, 2020