two persons a and b stand near a cliff and a fired a gun , b hears 2 sounds at an interval of 4sec . now b moves340 m towards a ... and hears 2 echoes at 6 sec interval. find the distance between a and b and the speed of sound
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Answer:
Distance of person from nearer cliff, (d
1
)
=
2
v×t
⇒d
1
=
2
330×3
=495m
Distance of person from farther cliff, d
2
=
2
v×t
d
2
=
2
330×5
=825m
Total distance between two cliffs
=d
1
+d
2
=(495+825)
=1320m.
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