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i have a problem with a question we have been given at school.

I have to find the distance (d) a bow-shooter is from the target, given the speed of the arrow (v1) and the speed of sound (v2), and the total time it took for the shooter to hear the arrow hitting the target after shooting it.

v1= 40 m/s v2= 390 m/s t1= 1 s

I have figured out that v1t2+v2t3 = 2d; t2+t3 = t

However, i am unable to put these two equations together to be able to solve this.

Thoughtfox1
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2 Answers2

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Note that you also have $v_1t_1=v_2t_2=d$. Use these two equations instead of the first one.

Botond
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Distance covered by the arrow is same as that covered by sound.
When distance covered is same then speed and time are inversely proportional to one another.

So, $v_1:v_2=4:39 \Rightarrow t_1:t_2=39:4$
Since, $t_1+t_2=1\Rightarrow39k+4k=1$
$\displaystyle \Rightarrow k=\frac1{43}$

Thus, distance, $d=v_1t_1=40\cdot\frac{39}{43}=36.28m$

InanimateBeing
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