You are in a hot-air balloon that, relative to the ground, has a velocity of 6.14 m/s in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is 2.44 m/s. What are (a) the magnitude and (b) direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.
To find the magnitude and direction of the hawk's velocity relative to the ground, we can use vector addition.
First, let's break down the velocities into their respective components. The velocity of the balloon relative to the ground (VBG) is 6.14 m/s due east, and the velocity of the hawk relative to the balloon (VHB) is 2.44 m/s due north.
Now we can add the two vectors together to find the velocity of the hawk relative to the ground (VHG). To do this, we add the x-components and y-components of the velocities separately.
In the x-direction:
VBGx = 6.14 m/s (east)
VHBx = 0 m/s (no horizontal component)
VHGx = VBGx + VHBx = 6.14 m/s (east)
In the y-direction:
VBGy = 0 m/s (no vertical component)
VHBy = 2.44 m/s (north)
VHGy = VBGy + VHBy = 2.44 m/s (north)
Now we have the x and y components of the hawk's velocity relative to the ground. To find the magnitude and direction, we can use the Pythagorean theorem and the arctangent function.
Magnitude (VHG):
VHG = √(VHGx^2 + VHGy^2)
VHG = √(6.14^2 + 2.44^2) m/s
VHG ≈ 6.61 m/s
Direction (θ):
θ = arctan(VHGy / VHGx)
θ = arctan(2.44 / 6.14)
Using a calculator or trigonometric table, we find:
θ ≈ 21.1 degrees
Therefore, the magnitude of the hawk's velocity relative to the ground is approximately 6.61 m/s, and its direction is 21.1 degrees north of east.