Question

In a TV set, an electron beam moves with horizontal velocity of 4.2 × 107 m/s across the cathode ray tube and strikes the screen, 41 cm away. The acceleration of gravity is 9.8 m/s How far does the electron beam fall while traversing this distance (in m)?

Answer 1

`sf(4.9xx10^(-18)color(white)(x)m)`

Explanation:

We can get the time of flight from the horizontal component, which is constant:

`sf(t=0.41/(4.2xx10^(-7))=9.762xx10^(-9)color(white)(x)s)`

Now we can consider the vertical component:

`sf(s_y=1/2"g"t^2)`

`sf(s_y=1/2xx9.8xx(9.762xx10^(-9))^2=4.862xx10^(-18)color(white)(x)m)`

`sf(s_y=4.9color(white)(x)"am")`

This must be an old question as I don't think there are many cathode ray t.v sets around anymore.