A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 3.6 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 60° when they move past the window, as the drawing shows. How fast is the train moving?

To solve this problem, we can use trigonometry and the concept of relative velocity.

Let's assume the speed of the train is v (in m/s), and the speed of the raindrops relative to the train is 3.6 m/s.

When the train is stationary, the raindrops fall vertically downwards, forming a right angle with the window. However, when the train starts moving, the raindrops appear to move at an angle of 60° with the vertical.

To find the speed of the train, we need to consider the relative motion between the train and the raindrops.

From the given information, we can see that the angle between the vertical direction and the direction of raindrops (60°) is formed by a right-angled triangle. The opposite side of the angle is the relative speed of raindrops to the train (3.6 m/s), and the hypotenuse is the total velocity of raindrops relative to the ground.

To find the total velocity of raindrops relative to the ground, we can use trigonometry. The opposite side represents the vertical component of the velocity and the hypotenuse represents the total velocity (relative to the ground).

We can use the equation:

sin(60°) = opposite/hypotenuse

sin(60°) = 3.6/v

Simplifying, we get:

√3/2 = 3.6/v

Cross-multiplying:

√3 * v = 2 * 3.6

v = (2 * 3.6) / √3

v ≈ 4.16 m/s

Therefore, the train is moving at approximately 4.16 m/s.

To solve this problem, we can use trigonometry. Let's assume that the speed of the train is v m/s relative to the ground.

When the train is stationary, the raindrops fall vertically and have a speed of 3.6 m/s relative to the ground, as mentioned in the problem.

When the train is moving, the raindrops appear to make an angle of 60° when they move past the window. We can consider this angle as the angle of elevation of the raindrops relative to the window.

Since the raindrops fall vertically (90°) relative to the ground when the train is stationary, the angle of elevation formed when the train is moving is the difference between the angle at which the raindrops are falling relative to the ground (90°) and the angle at which they are moving past the window (60°). So, the angle of elevation relative to the window is 90° - 60° = 30°.

Now, we can use trigonometry to find the relationship between the vertical speed of the raindrops (3.6 m/s) and the speed of the train (v m/s).

Using the trigonometric function tangent, we have the equation:
tan(30°) = (3.6 m/s) / v

Simplifying the equation, we have:
tan(30°) = 3.6 / v

Using a calculator, we can find that the tangent of 30° is roughly 0.577.

Substituting this value back into the equation, we have:
0.577 = 3.6 / v

To isolate v, we can multiply both sides of the equation by v:
0.577v = 3.6

Then, we can divide both sides of the equation by 0.577 to solve for v:
v = 3.6 / 0.577 ≈ 6.23

Therefore, the train is moving at a speed of approximately 6.23 m/s relative to the ground.