An anti-tank gun fires straight at a tank. The explosion of the shell is seen at the battery after $0.6\ \mathrm{s}$ and the sound of the explosion is heard $2.1\ \mathrm{s}$ after firing.
The velocity of sound may be taken as $340\ \mathrm{\displaystyle \frac{m}{s}}$
a) What is the distance from the battery to the tank?
$S=510\ \mathrm{m}$
Since the velocity of light is many times greater than the velocity of sound in air, the time $t_1$ at which the explosion was seen can be taken as the time of flight of the shell. Then the time $t_2$ at which the explosion was heard is the sum of the time of flight of the shell and the time it took for the sound to travel from the point of the explosion to the gun. Therefore the time taken by the sound to travel this distance is $t_2-t_1$ and the distance travelled by the shell is
$$S=v(t_2-t_1)$$
where $v$ is the velocity of sound.
b) What was the horizontal velocity of the projectile?
$u=850\ \mathrm{\displaystyle \frac{m}{s}}$
The velocity of the shell is
$$u=\frac{s}{\ t_1}$$
$$u=\frac{v(t_2-t_1)}{t_1}$$