Question

A boy stands in between two parallel cliffs and explodes a cracker. He hears the first echo after 0.5s and second echo 1.8s after listening to the first echo. Calculate the distance between the cliffs if speed of sound in air is 356 m s-1

Answer 1

Answer:

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Answer 2

Explanation:

Given A boy stands in between two parallel cliffs and explodes a cracker. He hears the first echo after 0.5 s and second echo 1.8 s after listening to the first echo. Calculate the distance between the cliffs if speed of sound in air is 356 m s-1

• Given a boy stands in between two parallel cliffs and bursts a cracker.
• So let P and Q be the two cliffs at a distance d1 and d2 meter from the boy.
• Given for d1, v = 356 m/s, t = 0.5 s
• So for d2, v = 356 m/s, t = 1.8 s
• So we have d1 = v x t / 2
•                         = 356 x 0.5 / 2
•                         = 178 / 2
•                           = 89 m
• Also we have d2 = v x t / 2
•                            = 356 x 1.8 / 2
•                            = 640.8 / 2
•                             = 320.4
• So distance PQ = d1 + d2
•                          = 89 + 320.4
•                          = 409.4 m

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https://brainly.in/question/1468067

https://brainly.in/question/1743681