A particle is projected with a velocity of `20 ms^(-1)` at an angle of `60^(@)` to the horizontal. The particle hits the horizontal plane again during its journey. What will be the time of impact ?

Options

A. `3.53 s`
B. `2.4s`
C. `1.7s`
D. 1s

Explanation

To find the time of impact for a particle projected with a velocity of \(20 \, \text{m/s}\) at an angle of \(60^\circ\) to the horizontal, we can use the formula for the time of flight in projectile motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity, \(u = 20 \, \text{m/s}\)
- Angle of projection, \(\theta = 60^\circ\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)

2. **Use the Time of Flight Formula:**
The time of flight \(T\) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]

3. **Calculate \(\sin \theta\):**
For \(\theta = 60^\circ\):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

4. **Substitute the Values into the Formula:**
Now substituting the values into the time of flight formula:
\[ T = \frac{2 \times 20 \times \sin 60^\circ}{9.8} \]
\[ T = \frac{2 \times 20 \times \frac{\sqrt{3}}{2}}{9.8} \]

5. **Simplify the Expression:**
The \(2\) in the numerator and denominator cancels out:
\[ T = \frac{20 \sqrt{3}}{9.8} \]

6. **Calculate the Numerical Value:**
Using \(\sqrt{3} \approx 1.732\):
\[ T \approx \frac{20 \times 1.732}{9.8} \approx \frac{34.64}{9.8} \approx 3.53 \, \text{seconds} \]

### Final Answer:
The time of impact is approximately \(3.53 \, \text{seconds}\).

---