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 Question from anu, a student: an electron in a TV tube is beamed horizontally at a speed of (50^6) m/sec. towards the face of a tube 40 cm away about how far will the electron drop before it hits? no information has been provided of initial height from where it is beamed.

Anu,

an electron in a TV tube is beamed horizontally at a
speed of (50^6) m/sec.

Do you mean $5 \times 10^6?$ Or $50 \times 10^6$ (engineering notation)? $50^6$ m/sec is about fifty times the speed of light!

towards the face of a tube 40 cm away
about how far will the electron drop before it hits?

Solve

$x - x_0 = v t$

(unaccelerated horizontal motion) to find time. Then plug into

$\Delta Y = \frac12 a t^2$ with $a = g \sim -9.8 \mbox{ m/sec}^2$

to find vertical displacement.

no information has been provided of initial height from where it is beamed.

So you can't find the final height, only how far it falls.

As you might expect, the answer you get should be effectively zero (well under a picometer, $10^{-12}$ meters) Gravity doesn't do much to an electron beam at this speed!

Good hunting!
RD

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.