We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:

which means

The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (

):

Therefore, the kinetic energy gained by the proton is

<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
a). (343 m/s) x (4.5 sec) = 1,543.5 meters
b). The surface it reflected from is (1543.5 / 2) = 771.75 meters away.
The sound had to go there AND come back before you heard the echo.