# Rain's velocity with respect to a moving car

A car travels due east with a horizontal speed of 44.4 km/h. Rain is falling vertically with respect to Earth. The traces of the rain on the side windows of the car make an angle of 68.4 degrees with the vertical.
(a) Find the velocity of the rain with respect to the car.

My attempt:
I thought perhaps I could model the solution from the problem involving the ground velocity of an airplane, as it flies through a wind that blows a different direction.

sin(theta)/V_rain = sin(90)/44.4 km/h

sin(68.4)/V_rain = sin(90)/44.4 km/h

44.4 km/h(0.930) = sin(90)(V_rain)
V_rain = 41.3 km/h

(b) Find the velocity of the rain with respect to Earth.
(Haven't attempted this one yet.)

All the Best,
Zem

Last edited:

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Päällikkö
Homework Helper
I'd say, with basic trigonometry, $$sin\theta = \frac{v_x}{v}$$, where $$v_x$$ is the horzontal speed of the rain (relative to the car), and v the "whole" speed.

b can be solved with tan (or alternatively, the Pythagorean theorem).

BobG
Homework Helper
Zem said:
A car travels due east with a horizontal speed of 44.4 km/h. Rain is falling vertically with respect to Earth. The traces of the rain on the side windows of the car make an angle of 68.4 degrees with the vertical.
(a) Find the velocity of the rain with respect to the car.

My attempt:
I thought perhaps I could model the solution from the problem involving the ground velocity of an airplane, as it flies through a wind that blows a different direction.

sin(theta)/V_rain = sin(90)/44.4 km/h

sin(68.4)/V_rain = sin(90)/44.4 km/h

44.4 km/h(0.930) = sin(90)(V_rain)
V_rain = 41.3 km/h

(b) Find the velocity of the rain with respect to Earth.
(Haven't attempted this one yet.)

All the Best,
Zem
Päällikkö has the right idea. You need to visualize what's happening.

There's a vertical component to the rain's velocity - directly towards the Earth. Relative to the Earth, there is no horizontal velocity to the rain.

There's a horizontal component to the car's velocity, but no vertical velocity.

Since your frame of reference in part a is the car, you have a vertical component and a horizontal component to the rain. The horizontal component is equal to the car's velocity (since the car is the frame of reference, the car appears stationary). The rain's linear speed can be found using the Pythagorean theorem and it has to be greater than either of its components.

In this case, you have the angle from the vertical and the horizontal component. Since you have the angle from the vertical, you use straight down as your reference point and rotate 68.4 degrees. You can work your way backwards via:

$$sin 68.4 = \frac{44.4}{V_{rain}}$$

Part b can be found using the the answer from part a and the cosine of 68.4 degrees. Or, if you want to stick to the original givens (always gives better accuracy), you can use the tangent.