A cyclist on the ground goes round a ciruclar path of circumference 34.3 m in `sqrt(22)` second. The angle made by him, with the vertical, will be:-

Options

A. `45^(@)`
B. `40^(@)`
C. `42^(@)`
D. `48^(@)`

Explanation

To solve the problem, we need to find the angle made by the cyclist with the vertical while moving in a circular path. We will follow these steps:

### Step 1: Calculate the Radius of the Circular Path
Given the circumference \( C = 34.3 \, \text{m} \), we can use the formula for the circumference of a circle:

\[ C = 2 \pi r \]

Rearranging for \( r \):

\[ r = \frac{C}{2 \pi} = \frac{34.3}{2 \pi} \]

Calculating the value:

\[ r \approx \frac{34.3}{6.2832} \approx 5.46 \, \text{m} \]

### Step 2: Calculate the Velocity of the Cyclist
The time taken to complete the circular path is given as \( t = \sqrt{22} \, \text{s} \). The velocity \( v \) can be calculated using the formula:

\[ v = \frac{\text{distance}}{\text{time}} = \frac{C}{t} = \frac{34.3}{\sqrt{22}} \]

Calculating the value:

\[ v \approx \frac{34.3}{4.690} \approx 7.31 \, \text{m/s} \]

### Step 3: Use the Formula to Find the Angle
We can use the relationship between the angle \( \theta \) and the other parameters. The formula is:

\[ \tan \theta = \frac{v^2}{rg} \]

Where \( g \approx 9.8 \, \text{m/s}^2 \) (acceleration due to gravity). Substituting the values we have:

\[ \tan \theta = \frac{(7.31)^2}{(5.46)(9.8)} \]

Calculating \( v^2 \):

\[ v^2 \approx 53.44 \]

Now substituting into the equation:

\[ \tan \theta = \frac{53.44}{(5.46)(9.8)} \approx \frac{53.44}{53.468} \approx 1 \]

### Step 4: Calculate the Angle
To find \( \theta \):

\[ \theta = \tan^{-1}(1) \]

This gives:

\[ \theta = 45^\circ \]

### Final Answer
The angle made by the cyclist with the vertical is \( 45^\circ \).

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