If a particle is projected with a velocity `49 m//"sec"`. making an angle `60^(@)` with the horizontal, its time of flight is given by

Options

A. `10sqrt(3)` sec
B. `5sqrt(3)` sec
C. `sqrt(3)` sec
D. None of these

Answer

To find the time of flight of a particle projected with a velocity of 49 m/s at an angle of 60 degrees with the horizontal, we can use the formula for time of flight in projectile motion:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity (u) = 49 m/s
- Angle of projection (θ) = 60 degrees
- Acceleration due to gravity (g) = 9.8 m/s²

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

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

4. **Substitute the Values into the Formula:**
Now, substituting the values into the time of flight formula:
\[ T = \frac{2 \times 49 \times \sin 60^\circ}{9.8} \]
\[ T = \frac{2 \times 49 \times \frac{\sqrt{3}}{2}}{9.8} \]

5. **Simplify the Expression:**
The 2 in the numerator and the 2 in the denominator cancel out:
\[ T = \frac{49 \sqrt{3}}{9.8} \]

6. **Calculate the Numerical Value:**
Now, simplifying further:
\[ T = \frac{49 \sqrt{3}}{9.8} = \frac{49 \sqrt{3}}{98/10} = \frac{49 \times 10 \sqrt{3}}{98} \]
\[ T = \frac{490 \sqrt{3}}{98} = 5 \sqrt{3} \text{ seconds} \]

### Final Answer:
The time of flight is \( 5 \sqrt{3} \) seconds.

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