A vector has a magnitude of 2m, but is pointing in an unknown direction. What is the maximum possible magnitude of its horizontal component?

1. Break Down the Problem

To find the maximum possible magnitude of the horizontal component of a vector with a given magnitude, we need to consider the relationship between the vector's magnitude and its components. The horizontal component can be defined based on the angle it makes with the horizontal axis.

2. Relevant Concepts

The horizontal component $V_{x}$ of a vector $V$ can be expressed using trigonometric functions as: $V_{x} = V \cdot \cos(\theta)$ where:

• $V$ is the magnitude of the vector (2 m in this case),
• $\theta$ is the angle between the vector and the horizontal axis.

To maximize $V_{x}$, we need $\cos(\theta)$ to be maximized.

3. Analysis and Detail

The maximum value of $\cos(\theta)$ is 1, which occurs when $\theta = 0^\circ$. Thus: $V_{x_{max}} = V \cdot \cos(0^\circ) = 2 \, \text{m} \cdot 1 = 2 \, \text{m}$

4. Verify and Summarize

We have determined that the maximum possible horizontal component occurs when the vector is directed entirely along the horizontal direction (0 degrees). Therefore, the horizontal component can indeed reach the full magnitude of the vector itself.

Final Answer

The maximum possible magnitude of the horizontal component is 2 m.