A bullet with an initial velocity of $c=99 \mathrm{msec}^{-1}$ travels vertically upward; its explosion is heard 5 sec after the start of its journey. At what height does the explosion occur, if the speed of sound is $c=333 \mathrm{msec}^{-1}$ and the acceleration due to gravity is $g=9,806 \mathrm{msec}^{-2} ? (0.001$ accuracy; neglect air resistance.)





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Ha a robbanást, a robbanás pillanatától számítva, $x$ sec múlva halljuk, akkor:

$$
99(5-x)-4,903(5-x)^{2}=333 x
$$

vagy rendezve:

$$
4,903 x+2+382,97 x-372,425=0
$$

mely egyenletnek positív gyöke:

$$
x=0,960636
$$

Így tehát a robbanás

$$
333 \cdot 0,960636=319,892 \mathrm{~m}
$$

magasságban történt.

If we hear the explosion $x$ seconds after the moment of the explosion, then:

$$
99(5-x)-4.903(5-x)^{2}=333 x
$$

or rearranged:

$$
4.903 x+2+382.97 x-372.425=0
$$

the positive root of this equation is:

$$
x=0.960636
$$

Thus, the explosion occurred at a height of

$$
333 \cdot 0.960636=319.892 \mathrm{~m}
$$