A man standing in front of a vertical cliff fires a gun. He hears the echo after 3.5 seconds. On moving closer to the cliff by 84 m, he hears the echo after 3 seconds. Calculate the distance of the cliff from the initial position of the man.

Given: v = speed of sound; first echo after t₁ = 3.5 s at distance d;

after moving 84 m closer, echo after t₂ = 3 s.

For an echo, 2 × distance = speed × time.

First position: 2d = v × 3.5 … (1) Second position: 2(d − 84) = v × 3 … (2)

Subtract (2) from (1): 2d − 2(d − 84) = v(3.5 − 3) 2d − 2d + 168 = 0.5v 168 = 0.5v v = 336 m s⁻¹

Substitute back into (1): 2d = 336 × 3.5 2d = 1 176 d = 588 m

Distance of the cliff from the initial position = 588 m

Two unknowns — the distance and the speed of sound — so you need two equations, one for each position. Write them both down before trying to solve anything.

Subtracting eliminates d in one move and hands you the speed. Students who try to guess the speed as 330 or 340 m s⁻¹ get a near-miss answer; the question wants you to derive it, and here it comes out at 336 m s⁻¹.

Keep the factor of 2 in BOTH equations. It is the same echo situation in each position, so if you drop it from one you will get a nonsense answer. And note the man moves CLOSER, so the second distance is d − 84, not d + 84.