Final answer:
An observer in a moving car would measure the raindrops to have a velocity magnitude of 21.21 m/s and falling at an angle of 8.13 degrees below the horizontal.
Step-by-step explanation:
To calculate the velocity of the raindrops as measured by an observer in a car moving at 21.0 m/s, we must consider the relative velocities of the raindrops and the car. Since the raindrops fall vertically at 3 m/s relative to the earth, and the car is moving horizontally at 21.0 m/s, we use vector addition to find the resultant velocity vector as seen by the observer in the car.
The vertical component of the raindrops' velocity remains at 3 m/s downward, and the horizontal component of the velocity is -21.0 m/s relative to the car (assuming the car's motion is in the positive direction). Therefore, the magnitude of the velocity vector can be found using the Pythagorean theorem:
Velocity magnitude = sqrt((vertical speed)^2 + (horizontal speed)^2)
Which becomes:
Velocity magnitude = sqrt((3 m/s)^2 + (21 m/s)^2) = sqrt(9 + 441) = sqrt(450) = 21.21 m/s
The direction of this velocity vector is given by:
tan(\( \theta \)) = vertical speed / horizontal speed
\( \theta \) = arctan(vertical speed / horizontal speed)
\( \theta \) = arctan(3 m/s / 21 m/s) = arctan(1/7) = 8.13 degrees below the horizontal
Thus, an observer in a moving car would measure the raindrops falling at an angle of 8.13 degrees below the horizontal, with a magnitude of 21.21 m/s.