A particle is projected at an angle of `45^(@)` with a velocity of `9.8 ms^(-1)`. The horizontal range will be (Take, `g = 9.8 ms^(-2))`

Options

A. `9.8m`
B. `4.9m`
C. `(9.8)/(sqrt(2))m`
D. `9.8 sqrt(2)m`

Answer

To find the horizontal range of a particle projected at an angle of \(45^\circ\) with an initial velocity of \(9.8 \, \text{m/s}\), we can use the formula for the range \(R\) of projectile motion:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

where:
- \(u\) is the initial velocity,
- \(\theta\) is the angle of projection,
- \(g\) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the parameters**:
- Initial velocity, \(u = 9.8 \, \text{m/s}\)
- Angle of projection, \(\theta = 45^\circ\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)

2. **Calculate \(\sin(2\theta)\)**:
- Since \(\theta = 45^\circ\),
- \(2\theta = 90^\circ\)
- Therefore, \(\sin(90^\circ) = 1\).

3. **Substitute the values into the range formula**:
\[ R = \frac{(9.8)^2 \cdot \sin(90^\circ)}{9.8} \]
\[ R = \frac{(9.8)^2 \cdot 1}{9.8} \]

4. **Simplify the expression**:
\[ R = \frac{9.8^2}{9.8} = 9.8 \, \text{m} \]

5. **Conclusion**:
The horizontal range \(R\) is \(9.8 \, \text{m}\).

### Final Answer:
The horizontal range will be \(9.8 \, \text{m}\).

---