A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 8.2 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 62° when they move past the window. How fast is the train moving?

Wouldn't tan 62= V/6.2

Of course, it is not clear how the angle is measured, either to the vertical or horizontal.

opps, typo, tan 62=V/8.2

15.42

To determine the speed of the train, we can use trigonometric relationships. Let's break down the problem:

Let v be the velocity of the train relative to the ground, and let θ be the angle the raindrops make with respect to the vertical when observed from the moving train.

From the perspective of the person in the stationary train, the motion of the raindrops appears vertical, so the vertical component of their velocity remains 8.2 m/s.

When the train moves, the motion of the raindrops consists of two components: the vertical component and the horizontal component due to the train's velocity.

Therefore, we need to resolve the velocity vector of the raindrops into horizontal and vertical components.

The vertical component of the raindrop's velocity is still 8.2 m/s, as this remains unchanged.

The horizontal component of the raindrop's velocity is v, the velocity of the train.

Using trigonometry, we can relate the vertical and horizontal components of the raindrop's velocity to the angle θ:

tan(θ) = vertical component / horizontal component
tan(θ) = 8.2 m/s / v

Rearranging this equation to solve for v, we have:

v = 8.2 m/s / tan(θ)

Plugging in the given angle θ of 62°, we can now calculate the speed of the train:

v = 8.2 m/s / tan(62°)

Calculating this expression, we find that the speed of the train is approximately 18.4 m/s.