A bat flying in a cave emits a sound and receives its echo 0.1 s later. Show that its distance from the cave wall is 19 m.

Solution 4P We can neglect the own speed of the bat, since it’s very small in comparison with the speed of sound. The sound wave emitted by the bat travels to the cave wall, where it’s reflected from the wall, and travels back to be received by the bat as an echo. Thus the sound wave travels twice the distance between the bat and cave wall. Therefore, the distance between the bat and the cave wall can be calculated as, L = wt 2 Where, w is the speed of the sound, t is the time of the sound wave travelling. We know that w = 19 m and t = 0.1 s in the above equation, = 19 m2× 0.1 s L = 0.95 m