Question

Got the ans thanks neede coz i had doubt but got clarified

Answer 1

Answer:

i. 170 m

ii. 850 m

Explanation:

Question

Arjun and his son Arav were standing ‘x’ m away from each other. They are equidistant (Y m) from a vertical cliff. Arjun burst a balloon and Arav heard the direct sound 0.5 seconds later and the echo after 2 seconds. If the speed of sound in air is 340m/s ,calculate

i. the distance between Arjun and Arav

ii. the distance between the cliff and Arjun

The parameters given are;

The time for the sound to reach Arav, $t_D$ = 0.5 s

Time for the echo to echo to reach Arav, $t_E$ = 2 s

The distance between Arjun and Arav = $D_D$

The distance between Arav and the cliff = $D_E$

The speed of sound in air, s = 340 m/s

The formula for speed, s, is $s = \dfrac{Distance, D}{Time, t}$, therefore;

$s = \dfrac{D_D}{t_D} = \dfrac{D_D}{0.5} = 340$

$D_D$ = 0.5 s × 340 m/s = 170 m

The distance between Arjun and Arav = 170 m

ii. Since Arav hears the direct sound before the echo, he is closer to the cliff than Arjun, therefore, we have;

$s = \dfrac{D_E}{t_E} = \dfrac{D_E}{2} = 340$

$D_E$ = 2 s × 340 m/s = 680 m

Therefore, the distance between Arav and the cliff = 680 m

Which gives the distance between the cliff and Arjun, $D_{cliff}$ = The distance between Arav and the cliff + The distance between Arjun and Arav

$D_{cliff}$ = 680 + 170 = 850 m.