



 
Anu,
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!
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.
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! Penny  


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