Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line measure as the velocity of the raindrops? (Assume the car is moving to the right and that the +x-axis is to the right. Enter the magnitude in m/s and the direction in degrees counterclockwise from the +x-axis.)

The magnitude of the resultant velocity is 21.55 m/s in the +x-axis and the direction is 69.61° counterclockwise from the +x-axis.

The magnitude of the resultant vectors of two perpendicular objects can be computed by using the formula:

[tex]\mathbf{|v_R| = \sqrt{(|v_1|)^2 + (|v_2|)^2}}[/tex]

From the given information;

let the velocity of the rain with respect to the ground be [tex]\mathbf{v_{rg}}[/tex]; and
the velocity of the car to the ground be = [tex]\mathbf{v_{cg}}[/tex]

∴

[tex]\mathbf{|v_R| = \sqrt{(|v_{rg}|)^2 + (|v_{cg}|)^2}}[/tex]

replacing the values, we have:

[tex]\mathbf{|v_R| = \sqrt{(7.5 \ m/s)^2 + (20.2 \ m/s)^2}}[/tex]

[tex]\mathbf{|v_R| = \sqrt{56.25 + 408.04}}[/tex]

[tex]\mathbf{|v_R| = \sqrt{464.29 \ m/s}}[/tex]

[tex]\mathbf{|v_R| =21.55 \ m/s}[/tex] in the +x-axis

The direction of the resultant velocity of the raindrop in counterclockwise direction i.e to the vertical direction is calculated by using the expression:

[tex]\mathbf{\theta = tan ^{-1} \Big |\dfrac{v_{cg}}{v_{rg}}\Big |}[/tex]

[tex]\mathbf{\theta = tan ^{-1} \Big |\dfrac{20.2}{7.5}\Big |}[/tex]

[tex]\mathbf{\theta = tan ^{-1} \Big |2.69\Big |}[/tex]

[tex]\mathbf{\theta = 69.61^0}[/tex]

Therefore, we can conclude that the magnitude of the resultant velocity is 21.55 m/s in the +x-axis and the direction is 69.61° counterclockwise from the +x-axis.

Learn more about the resultant vector here:

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