A car is moving at a constant 19.3m/s, and rain is falling at 8.9 m/s straight down. What angle does the rain make with respect to the horizontally observed by the driver?
To find the angle that the rain makes with respect to the horizontal as observed by the driver, we need to use trigonometry.
Let's denote the velocity of the car as Vc = 19.3 m/s and the velocity of the rain as Vr = 8.9 m/s.
The angle we are looking for is the angle between the horizontal direction and the direction of the rain as observed by the driver. Let's call this angle θ.
First, let's break down the velocity of the rain into its horizontal and vertical components. The vertical component of the rain's velocity is VRy = 8.9 m/s, as it is falling straight down. The horizontal component of the rain's velocity is VRx = 0 m/s since rain is not moving horizontally.
Now, let's consider the car's velocity. The car is moving horizontally, so its vertical component of velocity, VCy, is 0 m/s. The horizontal component of the car's velocity, VCx, is the same as the car's velocity, VCx = 19.3 m/s.
To find the angle θ, we can use the tangent function:
tan(θ) = VRy / VCx
Substituting the given values:
tan(θ) = 8.9 m/s / 19.3 m/s
Using a calculator or math software, you can find the inverse tangent (arctan) of this ratio to find the angle θ:
θ = arctan(8.9 / 19.3)
This should give you the angle in radians. If you want the answer in degrees, you can convert it by multiplying by (180 / π).
y velocity = 8.9
x velocity = 19.3
tan theta = (8.9/19.3)