We are tasked with determining how far south of its starting point the sparrow lands after completing its flight.

Given:

1. The sparrow flies 26 km at a bearing of 042° (42 degrees measured clockwise from North).
2. It then flies 49 km due south.

Approach:

• We first need to break the 26 km leg into two components: north-south (vertical) and east-west (horizontal) using basic trigonometry. The bearing of 42° allows us to use sine and cosine to resolve this.

  • The north-south distance (vertical component) is given by $26 \times \cos(42°)$.
  • The east-west distance (horizontal component) is given by $26 \times \sin(42°)$.

• Then, the total southward displacement will be the 49 km traveled due south minus the northward component from the first leg.

Let me calculate this for you.The sparrow lands 29.7 km south of its starting point.

Do you have any questions or would you like further details?

Here are 5 related questions to explore:

1. How would the answer change if the bearing were different, say 60°?
2. What would the total horizontal displacement be (east-west)?
3. How can trigonometry help in determining the exact landing point?
4. How would the answer differ if the sparrow flew north instead of south after the first leg?
5. What is the total distance traveled by the sparrow from start to finish?

Tip:

Always consider breaking down vector-based problems into their horizontal and vertical components for clarity.