Question

A cannon is fired at A (see diagram). An observer at B hears two sounds. The first sound is heard after 1 second and the second 3s after the observer sees the flash. (Velocity of sound = 340 ms⁻¹)
(i) Why does the observer see the flash before he hears the sound?
(ii) Calculate the distance x between the observer and the wall.

Answer 1

(i) An observer sees the flash first and hears the sound afterward because the velocity of light which is equal to 3 × 10⁸ ms⁻¹ is much higher than the velocity of sound which is 340 ms⁻¹.

(ii) v = `(3"d"_1)/"t"`

An echo or the second sound is heard after 3s.

∴ t = 3s

∴ 340 = `(3"d"_1)/3`

∴ d₁ = `(340xx3)/2`

= 510 ms⁻¹

In right Δ AOX: d₁² = OX² + d² = x² + d²

∴ d₁² − d² = x² ................(i)

but 2d = v × t .............(where t = 1s)

2d = 340 ..........(v = 340 ms⁻¹ and t = 1s)

∴ d = 170 m ....................(ii)

From equation (i) and (ii),

x = (510)² − (170)² = `10sqrt2312`

∴ x = 481 m

The distance between the observer and the well is 481 m.