Echo Ranger

Why it matters

Shout at a nearby wall: nothing. Back off past seventeen metres and the mountain suddenly answers. Your ear has a speed limit, and it's exactly one-tenth of a second.

The idea in plain words

Every sonar, every bat, one formula. Tap the terms.

d = v × t ÷ 2

A ship's sonar ping returns from the seabed in 1.2 s (v = 1500 m/s in water). Depth?

900 m

• metre: d — distance to the reflector
• m/s: v — speed of sound (≈ 344 m/s in air, ~1500 m/s in sea water)
• the classic trap: t ÷ 2 — the clock measures the ROUND trip — halve it, or your cliff is twice too far
• d = v × t ÷ 2 = 1500 × 1.2 ÷ 2

Predict first

To hear a distinct echo, the reflector must be at least ~17 m away because…

Persistence of hearing = 0.1 s. Distance = v×t/2 = 344 × 0.1 ÷ 2 ≈ 17 m. Closer than that, echo and original merge.

• the ear needs 0.1 s between sounds, and sound covers 34 m (there + back) in that time — correct
• sound is too weak closer than 17 m
• walls closer than 17 m don't reflect

What you do

Slide the cliff until the echo separates, then play sonar: place it so the echo returns in exactly half a second.

Check yourself

You hear your echo 0.6 s after shouting (v = 340 m/s). The cliff is…

d = 340 × 0.6 ÷ 2 = 102 m.

Bats catch insects in total darkness using…

They emit ultrasonic clicks and read the returning echoes — nature invented sonar first.

In a small classroom you never hear an echo because…

The reflection arrives within 0.1 s, so your brain hears one slightly-thickened sound, not two.

• 102 m away — correct
• 204 m away
• 51 m away
• ultrasonic echoes — sound ranging just like sonar — correct
• infrared vision
• smell
• the walls are nearer than 17 m — reflections merge with the original (reverberation at most) — correct
• classrooms absorb all sound
• voices are too quiet for echoes