Answer:
[tex]y=-4.9x10^{-16}m[/tex]
Explanation:
From the exercise we have initial velocity on the x-axis, the final x distance and acceleration of gravity.
[tex]v_{ox}=4.3x10^{7}m/s[/tex]
[tex]x=43cm=0.43m\\g=9.8m/s^{2}[/tex]
From the equation on moving particles we can find how long does it take the electron beam to strike the screen
[tex]x=x_{o}+v_{ox}t+\frac{1}{2}at^{2}[/tex]
Since [tex]x_{o}=0[/tex] and [tex]a_{x}=0[/tex]
[tex]0.43m=(4.3x10^{7}m/s)t[/tex]
Solving for t
[tex]t=1x10^{-8} s[/tex]
Now, from the equation of free-falling objects we can find how far does the electron beam fell
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
[tex]y=-\frac{1}{2}(9.8m/s^{2})(1x10^{-8} s)=-4.9x10^{-16}m[/tex]
The negative sign means that the electron beam fell from its initial point.
