# 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)?

Jan 17, 2018

$\textsf{4.9 \times {10}^{- 18} \textcolor{w h i t e}{x} m}$

#### Explanation:

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

$\textsf{t = \frac{0.41}{4.2 \times {10}^{- 7}} = 9.762 \times {10}^{- 9} \textcolor{w h i t e}{x} s}$

Now we can consider the vertical component:

$\textsf{{s}_{y} = \frac{1}{2} \text{g} {t}^{2}}$

$\textsf{{s}_{y} = \frac{1}{2} \times 9.8 \times {\left(9.762 \times {10}^{- 9}\right)}^{2} = 4.862 \times {10}^{- 18} \textcolor{w h i t e}{x} m}$

$\textsf{{s}_{y} = 4.9 \textcolor{w h i t e}{x} \text{am}}$

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