A man standing between two parallel hills fires a gun and hears two echoes, one 2.5 s and the other 3.5 s after the firing. If the velocity of sound is 330 ms-1, how long will it take him to hear the third echo?

The man is positioned between two parallel hills. The first two echoes represent the sound traveling to each hill and back. Let the distances to the hills be d1 and d2. The time for the first echo is t1 = 2d1/v = 2.5 s, and the second is t2 = 2d2/v = 3.5 s. The third echo occurs when the sound reflects off one hill, passes the man, reflects off the second hill, and then returns to the man [t5][t6]. The total time for this path is the sum of the time taken to reach the first hill and back (t1) plus the time taken to reach the second hill and back (t2), but since the third echo is a multiple reflection, the time is calculated as t1 + t2 [t5]. Thus, 2.5 s + 3.5 s = 6.0 s. This represents the sound completing one full round trip between the two hills [t4].