A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than the other. He fires a gun and, after a while, hears three echoes. The second echo arrives 1.05 s after the first, and the third echo arrives 0.827 s after the second. Assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance (in m) between the cliffs.

According to the given condition it can be assumed that the first echo would be heard from closest cliff. The second echo is from farther cliff and the third echo is from the reflection between the two cliffs.

Let the distance between the first cliff and the point of firing is x and y is the distance between the second cliff and the point of firing.

Then the first echo will travel 2x distance, second will travel 2y distance and third will travel 2x +2y.

So, using above data :

[tex]2y=v\times t_3[/tex]

and

[tex]2x=v(t_2+t_3)[/tex]

On solving :

[tex]y=\dfrac{343\times0.827}{2}[/tex]

[tex]y=141.8\ m[/tex]

[tex]x=\frac{343\times(1.05+0.827)}{2}[/tex]

[tex]x=321.9\ m[/tex]

x = 321.9 m and y = 141.8 m

Hence, total distance between two cliffs is d = 321.9 m + 141.8 m = 463.7 m