A person throws a bottle into a dust-bin at the same height as he is 2 m away at an angle of `45^(@)`. The velocity of the throw is

Options

A. g
B. `sqrt(g)`
C. 2g
D. `sqrt(2)g`

Answer

To solve the problem of a person throwing a bottle into a dust-bin at an angle of 45 degrees, we can follow these steps:

### Step 1: Understand the given information
- The horizontal distance (range) to the dust-bin, \( R = 2 \, \text{m} \)
- The angle of projection, \( \theta = 45^\circ \)
- The acceleration due to gravity, \( g \approx 9.81 \, \text{m/s}^2 \)

### Step 2: Use the range formula for projectile motion
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity of the throw.

### Step 3: Substitute the known values into the formula
Since \( \theta = 45^\circ \), we can calculate \( \sin(2\theta) \):
\[ \sin(2 \times 45^\circ) = \sin(90^\circ) = 1 \]
Now, substituting the values into the range formula:
\[ 2 = \frac{u^2 \cdot 1}{g} \]

### Step 4: Rearrange the equation to solve for \( u^2 \)
Multiplying both sides by \( g \):
\[ 2g = u^2 \]

### Step 5: Solve for \( u \)
Taking the square root of both sides:
\[ u = \sqrt{2g} \]

### Step 6: Substitute the value of \( g \)
Using \( g \approx 9.81 \, \text{m/s}^2 \):
\[ u = \sqrt{2 \times 9.81} \approx \sqrt{19.62} \approx 4.43 \, \text{m/s} \]

### Conclusion
The velocity of the throw is approximately \( 4.43 \, \text{m/s} \).

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