Question

A man stands in front of a cliff and fires a gun. He hears an echo after 4 second on moving closer to by 41.2 m again firing a gun he hears an echo after 3.75 s the distance of the man from the cliff to his initial position is and speed from sound​

Answer 1

{some corrections in question}

Correct question : A man stands in front of a vertical cliff and fires a gun. He hears an echo after 4 sec. On moving closer to the cliff by 41.2 m he fires again and hear the echo after 3.75 sec. Find the distance of the man from the cliff to his initial position is and speed the speed sound ?

Solution : A/C to question; Let, the distance of man from the cliff to his initial position be x. Due to echo, distance traveled by sound in 4 sec is 2x.

Now, by formula of speed,

⟹ speed = distance/time

⟹ s = 2x/4 ----{1}

Now, a/c to question, On moving closer to the cliff by 41.2 m he fires again and hear the echo after 3.75 m. Hence distance is x - 41.2. Due to echo, distance traveled by sound in 3.75 sec is 2( x - 41.2)

⟹ speed = distance/time

⟹ s = 2(x - 41.2)/3.75

⟹ s = 2x - 82.4 /3.75 ----{2}

From eq. {1} and eq. {2}

⟹ 2x/4 = 2x - 82.4 /3.75

⟹ 3.75 * 2x = 4 (2x - 82.4)

⟹ 7.5x = 8x - 329.6

⟹ 7.5x - 8x = - 329.6

⟹ - 0.5x = - 329.6

⟹ 0.5x = 329.6

⟹ x = 329.6/0.5

⟹ x = 659.2 m

Hence, the distance of man from the cliff to his initial position is 659.2 m.

Now, put value of x in eq. {1}

⟹ s = 2x/4

⟹ s = 2 * 659.2/4

⟹ s = 1318.4/4

⟹ s = 329.6 m/s

Hence, speed of the sound is 329.6 m/s.

Answer : Hence, the distance of man from the cliff to his initial position is 659.2 m and speed of the sound is 329.6 m/s.

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