# Angle rain appears to driver

1. Feb 4, 2013

### burton95

1. The problem statement, all variables and given/known data

Rain is falling vertically at a constant speed of 8 m/s. At what angle (in degrees) from the vertical will the rain appear to be falling as viewed by a driver travelling on a straight, level road with a speed of 50 km/h?

2. Relevant equations

vr,c = vr,g + vc,g

3. The attempt at a solution

I set vr,g = -8, and vc,g = 13.88 m/s after conversion. I came up with a 13.88 sinθ - 8 = 0. θ = 35.19. 35.19+90 = 125.19. 180 - 125.19 = 54.8 degrees off of vertical. Not even close.

2. Feb 4, 2013

### haruspex

By what reasoning do you arrive at that?

3. Feb 5, 2013

### burton95

My book had a similar problem. The reasoning was to get the another angle besides the 90° and then i could subtract the difference

4. Feb 5, 2013

From 180°

5. Feb 5, 2013

### HallsofIvy

Staff Emeritus
Essentially you have a right triangle with one leg of length and the other of length 13.8. tangent= opposite side/near side.

6. Feb 5, 2013

### burton95

Right I get that but I have a negative number as per my definition of up being positive. I understand -1tan = 13.8/8 but I'm at a loss as to how this would work as I assigned it a value of -8.....wait do I just use 8 as the magnitude and "-" is the direction so I only need the magnitude of the velcoity?????

7. Feb 5, 2013

### haruspex

You can do it either way - just work with magnitudes and figure out the direction separately, or understand that the tangent function goes negative in the second and fourth quadrants (tan(x) = - tan(π-x)).

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