in the figure given below, there are two boys, A and B
standing on the two peaks of hill of same height
and 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
INeglect height of A and B]
....
Answers
Hi there,
The figure for the question is missing. I have attached the figure below and have solved accordingly. Thanks.
Answer: h = [¼ * √{v² + 4sv}]
Explanation:
Let the height of peaks of the hill be “h” and velocity of the sound be “v”.
Distance between the two peaks is given as = s
When A shouts, B hears the sound at an interval of 0.5 s.
Here the sound will travel in two different ways:
Case (i): firstly, the first sound that B hears at t sec when A shouts.
Here, Time, t = distance/speed = s/v …. (i)
Case (ii): secondly, the echo of the sound that reaches to B after reflecting from the ground.
Referring to the figure below, since the distance between PQ is given as “s” so, PO = OQ = s/2.
Assuming that the reflection takes place from the middle of the ground, therefore, by using the Pythagoras theorem, we get
The distance covered by the echo of the sound AOB,
s’ = 2 * √[h² + (s/2)²] …. (ii)
Let the time that the echo will take be denoted as “ t’ ” = s’/v ……. (iii)
Thus,
According to the question, we have
t’ – t = 0.5
⇒ s’/v – s/v = 0.5 ….. [from (i) & (iii)]
substituting the value of s’ from (ii), we get
⇒ [2 * √{h² + (s/2)²}]/v – s/v = 0.5
⇒ [2 * √{h² + (s²/4}]/v = ½ + s/v
⇒ [2 * √{h² + (s²/4}]/v = [v+2s]/2v
⇒ 2 * [2 * √{h² + (s²/4)] = [v+2s]
squaring both sides
⇒ 16 [h² + (s²/4)] = v² + 4sv + 4s²
⇒ 16h² + 4s² = v² + 4sv + 4s²
⇒ 16h² = v² + 4sv
⇒ h² = 1/16 * [v² + 4sv]
taking square root on both sides
⇒ h = ¼ * √[v² + 4sv]
Thus, the height of the peaks is [¼ * √{v²+ 4sv}].
Hope this is helpful!!!!
