Respuesta :
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, [tex]t_D[/tex] = 0.5 s
Time for the echo to echo to reach Arav, [tex]t_E[/tex] = 2 s
The distance between Arjun and Arav = [tex]D_D[/tex]
The distance between Arav and the cliff = [tex]D_E[/tex]
The speed of sound in air, s = 340 m/s
The formula for speed, s, is [tex]s = \dfrac{Distance, D}{Time, t}[/tex], therefore;
[tex]s = \dfrac{D_D}{t_D} = \dfrac{D_D}{0.5} = 340[/tex]
[tex]D_D[/tex] = 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;
[tex]s = \dfrac{D_E}{t_E} = \dfrac{D_E}{2} = 340[/tex]
[tex]D_E[/tex] = 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, [tex]D_{cliff}[/tex] = The distance between Arav and the cliff + The distance between Arjun and Arav
[tex]D_{cliff}[/tex] = 680 + 170 = 850 m.
