In $\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{h}$, a balloon drifts $\mathbf{21}\mathbf{.}\mathbf{5}\mathit{k}\mathit{m}$north, $\mathbf{9}\mathbf{.}\mathbf{70}\mathbf{km}$east, and $\mathbf{2}\mathbf{.}\mathbf{88}\mathbf{km}$upward from its release point on the ground. Find (a) the magnitude of its average velocity. (b) the angle its average velocity makes with the horizontal.

1) Time the balloon travels for is, $\mathrm{t}=3.5\mathrm{h}$

2) Displacement of Balloon is, $21.5\mathrm{km}$ along North, $9.7\mathrm{km}$ along East and $2.88\mathrm{km}$ along upward direction.

Using the equation of average velocity, we can calculate the average velocity in each direction. Then, we can find the magnitude of average velocity by using those components. To calculate the angle, average velocity makes with the horizontal direction, we can use the equation of tangent.

Formula:

$v=\frac{d}{t}$

$\theta ={\tan }^{-1}\left(\frac{y}{X}\right)$

Suppose, x axis is along east, y-axis is along north and z-axis is along upward direction.

We have,

$v=\frac{d}{t}$

So, velocity along the x-axis or east is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{x}}=\frac{9.70\mathrm{km}}{3.50\mathrm{h}} \\ =2.77\mathrm{km}/\mathrm{h} \end{gathered}$$

The velocity along y-axis or north is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{y}}=\frac{21.5\mathrm{km}}{3.50\mathrm{s}} \\ =6.14\mathrm{km}/\mathrm{h} \end{gathered}$$

And, the velocity along z-axis or upward direction is,

$$\begin{gathered} {\mathrm{V}}_z=\frac{2.88\mathrm{km}}{3.50\mathrm{s}} \\ =0.82\mathrm{km}/\mathrm{h} \end{gathered}$$

So, the average velocity,

$$\begin{gathered} V=\sqrt{{\mathrm{V}}_{\mathrm{x}}^2+{\mathrm{V}}_{\mathrm{y}}^2+{\mathrm{V}}_{\mathrm{z}}^2} \\ =\sqrt{{\left(2.77\mathrm{km}/\mathrm{h}\right)}^2+{\left(6.14\mathrm{km}/\mathrm{h}\right)}^2+{\left(0.82\mathrm{km}/\mathrm{h}\right)}^2} \\ =6.79\mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore, the magnitude of the average velocity of the balloon is $6.79\mathrm{km}/\mathrm{h}$.

The angle can be calculated as,

$\theta ={\tan }^{-1}\left(\frac{y}{X}\right)$

Here, x is the vector in XY plane, it can be calculated as using the displacements along x and y axis. Therefore,

$$\begin{gathered} \mathrm{x}=\sqrt{{\left(9.7\mathrm{km}\right)}^2+{\left(21.5\mathrm{km}\right)}^2} \\ =23.58\mathrm{km} \end{gathered}$$

Hence,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{y}{X}\right) \\ ={\tan }^{-1}\left(\frac{2.88km}{23.58km}\right) \\ =6.96° \end{gathered}$$

Therefore, the angle that average velocity makes with the horizontal is $6.96°$.