Question

A balloon is rising at 3. 6 m/s and at the same time is being blown by wind of 5. 4 m/s from the east. What is the actual magnitude and direction of the velocity of the balloon?.

Answer 1

The magnitude of the balloon's velocity is 6.49 m/s, and its direction is 33.7 degrees above the horizontal (eastward direction). This is determined using the Pythagorean theorem for magnitude and trigonometry for direction.

Step-by-step explanation:

The question involves calculating the resultant velocity of a balloon that is rising vertically while also being pushed horizontally by the wind. To find the magnitude and direction of the balloon's actual velocity, we can use the Pythagorean theorem and trigonometry.

Calculating the magnitude:
We have two perpendicular velocity components: 3.6 m/s (vertical) and 5.4 m/s (horizontal). The magnitude of the resultant velocity (v) can be calculated using the Pythagorean theorem:

v = √(3.62 + 5.42)
v = √(12.96 + 29.16)
v = √(42.12)
v = 6.49 m/s

Calculating the direction:
The direction can be calculated using the tangent function:
θ = tan-1(vertical component / horizontal component)
θ = tan-1(3.6 / 5.4)
θ = tan-1(0.6667)
θ = 33.7°

The direction is 33.7 degrees above the horizontal (eastward direction) since this is the angle the vertical rise makes with the horizontal wind direction.

Answer 2

The actual magnitude of the velocity of the balloon is
`\(6.0 \, \text{m/s}\)`at an angle of
`\(53.1^\circ\)` north of east.

Step-by-step explanation:

To find the actual velocity of the balloon, we use vector addition because the balloon has both vertical and horizontal components of velocity. The vertical component is the rising speed of
`\(3.6 \, \text{m/s}\)`, and the horizontal component is the wind speed from the east of
`\(5.4 \, \text{m/s}\).`

Let
`\(v_{\text{vertical}}\)` be the vertical component, and
`v_{\text{horizontal}}\)` be the horizontal component.

`\[ v_{\text{vertical}} = 3.6 \, \text{m/s} \]`

`\[ v_{\text{horizontal}} = 5.4 \, \text{m/s} \]`

The magnitude of the actual velocity is given by the Pythagorean theorem:

`\[ v_{\text{actual}} = \sqrt{v_{\text{vertical}}^2 + v_{\text{horizontal}}^2} \]`

`\[ v_{\text{actual}} = \sqrt{(3.6 \, \text{m/s})^2 + (5.4 \, \text{m/s})^2} \]`

`\[ v_{\text{actual}} = √(12.96 + 29.16) \]`

`\[ v_{\text{actual}} = √(42.12) \]`

`\[ v_{\text{actual}} \approx 6.0 \, \text{m/s} \]`

The direction of the velocity is given by the arctangent:

`\[ \theta = \tan^(-1)\left(\frac{v_{\text{vertical}}}{v_{\text{horizontal}}}\right) \]`

`\[ \theta = \tan^(-1)\left((3.6)/(5.4)\right) \]`

`\[ \theta \approx \tan^(-1)(0.6667) \]`

`\[ \theta \approx 33.1^\circ \]`

The direction is measured north of east, so the actual velocity is
`\(6.0 \, \text{m/s}\)` at an angle of
`\(53.1^\circ\)` north of east.