Question

In the figure given below, there are two boys, A and B
standing on the two peaks of hill of same height
and s is the distance between peaks. A shouts and
B hears two sounds at an interval of 0.5 s. If speed
of sound in air is v, then height of peaks is
[Neglect height of A and B]
)
+Options
(1) sv
(2) 1 1 2 + sv-8)
co Tv2 + 4sv]"?​

Answer 1

Answer:

Option (3) $\frac{1}{4}\sqrt{v^2+4sv}$

Explanation:

If the first sound is heard by B after t seconds then

$t=\frac{s}{v}$

We will assume that the second sound wave reaches to A after refleacting from the middle of the ground

The second sound wave will be reflected from the ground and will reach to B

The distance covered

$s'=2\sqrt{h^2+(s/2)^2}$

Where h is the height of the cliffs

$\implies s'=2\sqrt{h^2+s^2/4}$

Time taken by this sound wave

$t'=\frac{s'}{v}$

$t'-t=0.5$

$\implies \frac{s'}{v}-\frac{s}{v}=0.5$

$\implies \frac{2\sqrt{h^2+s^2/4}}{v}-\frac{s}{v}=\frac{1}{2}$

$\implies \frac{2\sqrt{h^2+s^2/4}}{v}=\frac{1}{2}+\frac{s}{v}$

$\implies \frac{2\sqrt{h^2+s^2/4}}{v}=\frac{v+2s}{2v}$

$\implies 4\sqrt{h^2+s^2/4}=v+2s$

$\implies 16(h^2+s^2/4)=v^2+4s^2+4sv$

$\implies 16h^2+4s^2=v^2+4s^2+4sv$

$\implies 16h^2=v^2+8vs$

$\implies h^2=\frac{v^2+4sv}{16}$

$\implies h=\frac{1}{4}\sqrt{v^2+4sv}$

Hope this helps.