The shooter

The shooter heard the impact of the bullet on the target in one second after the shot. The bullet was moving at an average speed of 500 m/s. Calculate with speed of sound of 340 m/s. Determine the distance of the target.

Correct result:

x =  202.381 m

Solution:

$$\begin{gathered} t=1\text{ s } \\ {v}_1=500\text{ m/s } \\ {v}_2=340\text{ m/s } \\ {t}_1+t_2=t \\ x\mathrm{/}{v}_1+x\mathrm{/}{v}_2=t \\ x\mathrm{/}500+x\mathrm{/}340=1 \\ 42x=8500 \\ x={\displaystyle \frac{4250}{21}}\doteq 202.380952=202.381\text{ m} \end{gathered}$$

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