A falcon and its trainer are exactly midway between two vertical de parallel hills. As the res the
gun, the falcon starts flying directly towards one of the hists. The falcon hears first and for thoes of
the gun-shot at 2 s and 3 s respectively. Ignore reflection of sound from the falcon and from the trainer.
The air is practically still. Among the following options, respective speeds of the falcon ancitra sound (in
m/s) could be
Answers
Given info : As the trainer fires the gun, the falcon starts flying directly towards one of the hills. The falcon hears first and 2nd echos of the gun-shot at 2 s and 3 s respectively.
To find : respective speed of the falcon and sound could be..
solution : Let A and B are two hills and distance between AB is d. both falcon and trainer are located in midway of hills.
case 1 : for the first echo,
let speed of falcon is v and speed of sound is Vs .
falcon moves towards hills.
total distance covered by echo to reach falcon = d/2 + d/2 - v × 2
[ as first echo hears at 2 s ]
= d - 2v
∴ speed of sound , Vs = distance travelled by echo to reach falcon /time taken
= (d - 2v)/2
case 2 : for the 2nd echo
this time echo heard by falcon is come from the other hills.
so, distance covered = d/2 + d/2 + v × 3 [ 2nd echo hears at 3 s]
= d + 3v
∴ speed of sound, Vs = (d + 3v)/3
but speed of sound remains same.
so, (d - 2v)/2 = (d + 3v)/3
⇒3d - 6v = 2d + 6v
⇒d = 12v
then, Vs = (12v - 2v)/2 = 5v
Therefore speed of sound is five times of speed of falcon.
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Answer:
Given info : As the trainer fires the gun, the falcon starts flying directly towards one of the hills. The falcon hears first and 2nd echos of the gun-shot at 2 s and 3 s respectively.
To find : respective speed of the falcon and sound could be..
solution : Let A and B are two hills and distance between AB is d. both falcon and trainer are located in midway of hills.
case 1 : for the first echo,
let speed of falcon is v and speed of sound is Vs .
falcon moves towards hills.
total distance covered by echo to reach falcon = d/2 + d/2 - v × 2
[ as first echo hears at 2 s ]
= d - 2v
∴ speed of sound , Vs = distance travelled by echo to reach falcon /time taken
= (d - 2v)/2
case 2 : for the 2nd echo
this time echo heard by falcon is come from the other hills.
so, distance covered = d/2 + d/2 + v × 3 [ 2nd echo hears at 3 s]
= d + 3v
∴ speed of sound, Vs = (d + 3v)/3
but speed of sound remains same.
so, (d - 2v)/2 = (d + 3v)/3
⇒3d - 6v = 2d + 6v
⇒d = 12v
then, Vs = (12v - 2v)/2 = 5v
Therefore speed of sound is five times of speed of falcon.
Explanation: