Question

The angle of elevation of an airborne helicopter from a point A on the ground is 45°. After a flight of 15 seconds, the angle of elevation of the helicopter changes to 30°. If the helicopter is flying at a constant height of 2000 m, find the speed of the helicopter. (Take `sqrt(3) = 1.732`)

Answer 1

Given:

Angle of elevation at A initially = 45°.

After 15 s angle of elevation = 30°.

Constant height of helicopter h = 2000 m.

Use `sqrt(3) = 1.732`.

Step-wise calculation:

1. Let horizontal distances from A to the helicopter's ground-projection at the two instants be x₁ and x₂.

2. From `tan 45^circ = h/x_1` 

⇒ `1 = 2000/x_1` 

⇒ x₁ = 2000 m

3. From `tan 30^circ = h/x_2` 

⇒ `(1/sqrt(3)) = 2000/x_2` 

⇒ `x_2 = 2000 xx sqrt(3)`

= 2000 × 1.732

= 3464 m

4. Horizontal distance traveled in 15 s = Δ_x

= x₂ – x₁

= 3464 – 2000 

= 1464 m

5. Speed = `"Distance"/"Time"` 

= `(1464  m)/(15  s)` 

= 97.6 m/s

6. Convert to km/h if desired:

97.6 × 3.6 = 351.36 km/h

Speed of the helicopter = 97.6 m/s ≈ 351.36 km/h.