Question

You are riding in a hot air balloon that, relative to the ground, has a velocity of 6.0 m/s due east. You see a hawk moving directly away from the balloon with a velocity relative to you of 2.0 m/s due north. What are the magnitude and direction of the hawk’s velocity relative to the ground? Express the directional angle relative to due east. Note that both the balloon and the hawk are in horizontal motion.

Answer 1

Answer:Velocity = 6.325m/s

Directional angle= 18.43°

Explanation:

Using Right angle triangle

Let Velocity of ballon&hawk be VHB represent the height of the triangle.

Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.

Let Velocity of hawk and ground BE VHG represent the hypothesis.

Theta = opp/Adj= VHB/VBG

using pythagorean

VHG= SQRT(VHB^2+VBG^2)

VHG= sqrt(2^2+6^2)

VHG= sqrt(4+36)

VHG= 6.325m/s

Tan theta= 2/6

Tan theta =0.3333

Tan^-1 0.3333=18.43°