A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively the distance between the cliffs is: (Velocity of sound in air =340ms-1 )

Question

A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively the distance between the cliffs is: (Velocity of sound in air =340ms-1 ) (A) 190 m (B) 850 m (C) 591 m (D) 510 m

Tardigrade Admin · Accepted Answer

Let the person is at point P. Time taken in first echo,

t_{1} = 1.5 s

∴

x = \frac{V \times t _{1}}{2}

= \frac{340 \times 7.5}{2} = 255 m

Similarly,

y = \frac{V \times t _{2}}{2}

= \frac{340 \times 3.5}{2}

= 595 m

Distance between two cliffs

= x + y

= 255 + 595 = 850 m

Q. A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively the distance between the cliffs is : (Velocity of sound in air =340ms−1 )

190 m

850 m

591 m

510 m

Let the person is at point P. Time taken in first echo, t1​=1.5s ∴ x=2V×t1​​ =2340×7.5​=255m Similarly, y=2V×t2​​ =2340×3.5​ =595m Distance between two cliffs =x+y = 255 + 595 = 850 m

Q. A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively the distance between the cliffs is : (Velocity of sound in air $= 340 m s^{- 1}$ )

Let the person is at point P. Time taken in first echo, $t_{1} = 1.5 s$
$∴$ $x = \frac{V \times t _{1}}{2}$ $= \frac{340 \times 7.5}{2} = 255 m$ Similarly, $y = \frac{V \times t _{2}}{2}$ $= \frac{340 \times 3.5}{2}$ $= 595 m$ Distance between two cliffs $= x + y$ = 255 + 595 = 850 m