The parachutist's total velocity, $\vec{v}$, is the vector sum of the vertical velocity component, $\vec{v}_y$, and the horizontal velocity component, $\vec{v}_x$. In still air, the landing speed is purely vertical, so $v_y = 4.0$ m/s (downward). The wind provides a horizontal velocity component, $v_x = 2.0$ m/s (eastward). Since the vertical and horizontal components are perpendicular, we can use vector addition principles.

Let's define a coordinate system where the y-axis is downward and the x-axis is eastward. The velocity components are:

$\vec{v}_y$ with magnitude $v_y = 4.0$ m/s $\vec{v}_x$ with magnitude $v_x = 2.0$ m/s

The total velocity vector is $\vec{v} = \vec{v}_x + \vec{v}_y$.

[Q1] What is the parachutist's landing speed? The landing speed is the magnitude of the total velocity vector, $v = |\vec{v}|$. Since the components are orthogonal, we use the Pythagorean theorem.

Derivation:

$$v = \sqrt{v_x^2 + v_y^2}$$

Substitution:

$$v = \sqrt{(2.0 \text{ m/s})^2 + (4.0 \text{ m/s})^2}$$ $$v = \sqrt{4.0 + 16.0} \text{ m/s} = \sqrt{20.0} \text{ m/s}$$ $$v \approx 4.5 \text{ m/s}$$

[Q2] What is the direction of the landing velocity? The direction can be described by the angle $\theta$ that the total velocity vector makes with the downward vertical.

Derivation: From the vector components forming a right triangle, we can use the tangent function.

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{v_x}{v_y}$$ $$\theta = \arctan\left(\frac{v_x}{v_y}\right)$$

Substitution:

$$\theta = \arctan\left(\frac{2.0 \text{ m/s}}{4.0 \text{ m/s}}\right) = \arctan(0.50)$$ $$\theta \approx 27^\circ$$

This angle is measured from the downward vertical, towards the east.