An aeroplane taked off at an angle of `30^(@)` to the horizontal. If the component of its velocity along the horizontal is `200 kmh^(-1)`, what is its actual velocity ? Also find the vertical component of its velocity.

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the problem
The aeroplane takes off at an angle of \(30^\circ\) to the horizontal. We know the horizontal component of its velocity is \(200 \, \text{km/h}\). We need to find the actual velocity of the aeroplane and the vertical component of its velocity.

### Step 2: Use the relationship between components and actual velocity
The horizontal component of the velocity (\(V_x\)) can be expressed in terms of the actual velocity (\(V\)) and the angle (\(\theta\)):
\[ V_x = V \cos(\theta) \]
Given that \(V_x = 200 \, \text{km/h}\) and \(\theta = 30^\circ\), we can rearrange this formula to find the actual velocity (\(V\)):
\[ V = \frac{V_x}{\cos(\theta)} \]

### Step 3: Calculate the actual velocity
Substituting the known values into the equation:
\[ V = \frac{200 \, \text{km/h}}{\cos(30^\circ)} \]
Using the value of \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\[ V = \frac{200 \, \text{km/h}}{\frac{\sqrt{3}}{2}} = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, \text{km/h} \]

### Step 4: Find the vertical component of the velocity
The vertical component of the velocity (\(V_y\)) can be expressed as:
\[ V_y = V \sin(\theta) \]
Substituting the actual velocity we found and \(\sin(30^\circ) = \frac{1}{2}\):
\[ V_y = \left(\frac{400}{\sqrt{3}} \, \text{km/h}\right) \sin(30^\circ) = \left(\frac{400}{\sqrt{3}} \, \text{km/h}\right) \times \frac{1}{2} = \frac{200}{\sqrt{3}} \, \text{km/h} \]

### Final Answers
- Actual velocity of the aeroplane: \(\frac{400}{\sqrt{3}} \, \text{km/h}\)
- Vertical component of the velocity: \(\frac{200}{\sqrt{3}} \, \text{km/h}\)

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