Question

An observer X is on an island 5500 m from a vertical cliff on the shore. A ship Y is anchored between the island and the Cliff. A blast on the ship’s siren is heard twice by X, the interval being 5s. [Velocity of sound = 330 ms⁻¹]

(i) Find the distance of the ship from the island.
(i) Find the distance of the cliff from the ship.

Answer 1

(i) Let the distance between the observer X on the island and the ship Y = x m
Distance between the ship Y and the cliff = (5500 – x) m

Let t₁ be the time at which the first blast is heard.

∴ t₁ = `"x"/"v"`

= `"x"/330` ............`("v"="Distance"/"Time")` ...........(i)

Let t₂ be the time at which the second blast is heard i.e., time taken for the sound to travel from the ship to the cliff and then back to the observer on the island.

Total distance = Y to cliff + Cliff to the island

Total distance = (5500 − x) + 5500

= (11000 − x)

∴ t₂ = `"Total distance travelled"/"Velocity"`

∴ t₂ = `(11000-"x")/330` ...............(ii)

but t₂ − t₁ = 5s

∴ Subtracting equation (i) from equation (ii) we get

`(11000-"x")/330-"x"/300` = 5

∴ 11000 − 2x = 5 × 330

∴ 2x = 1100 − 1650

= 9350

x = 4675 m.

∴ Distance between the ship Y and the island = 4675 m.

(ii) Distance between ship y and the cliff

= 5500 − x

= 5500 − 4675

∴ x = 825.

The distance of the ship from the island is 4675 m and the distance of the cliff from the ship is 825m.