# A person looking out the window of a stationary train notices that raindrops are falling...

## Question:

A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 8.1 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of ? = 25 when they move past the window as the drawing shows. How fast is the train moving?

## vector components

In this problem, we are asked to determine the speed of the train, which is also equal to the horizontal speed of the rain ({eq}v_x{/eq}). To determine {eq}v_x{/eq} we used the following equation: {eq}\theta=tan^{-1}\frac{v_y}{v_x}{/eq}, where {eq}v_y{/eq} is the vertical speed of the rain.

Let the

vertical speed of the rain = {eq}v_y=8.1\ m/s{/eq}

horizontal speed of the rain = {eq}v_x{/eq}

direction of the rain = {eq}\theta=25^\circ{/eq}

The horizontal speed of the rain is equal to the speed of the train.

The direction of the rain is related to the vertical and horizontal speeds of the rain as \times10^{}ressed below:

$$\tan\theta=\frac{v_y}{v_x}$$

Solving for {eq}v_x:{/eq}

$$v_x=\frac{v_y}{\tan\theta}$$

Substituting the given values:

$$v_x=\frac{8.1\ m/s}{\tan{25^\circ}}=17.37\ m/s$$ 