A particle is projected at an angle $30^{\circ}$ to the horizon with a velocity of 1962 cm/sec. The time of flight is

Question

Accepted Answer

To determine the time of flight for a particle projected at an angle of $30^{\circ}$ with a velocity of 1962 cm/sec, we can use the formula for the time of flight $T$ of a projectile:

$$ T = \frac{2u \sin \theta}{g} $$

where:
- $u$ is the initial velocity (1962 cm/sec),
- $\theta$ is the angle of projection ($30^{\circ}$),
- $g$ is the acceleration due to gravity (approximately 980 cm/sec$^2$ for calculations in cm/sec).

Substituting the given values into the formula:

$$ T = \frac{2 \times 1962 \times \sin 30^{\circ}}{980} $$

Since $\sin 30^{\circ} = 0.5$, the equation becomes:

$$ T = \frac{2 \times 1962 \times 0.5}{980} $$

$$ T = \frac{1962}{980} $$

$$ T = 2 \text{ seconds} $$

Thus, the time of flight is 2 seconds, which corresponds to Option 2.

**Explanation for Incorrect Options:**

1. **Option 1 (1 sec):** This option is incorrect because, as calculated, the time of flight is 2 seconds. A time of 1 second would imply a much higher gravitational acceleration or a lower initial velocity, neither of which is applicable here.

2. **Option 3 (2.5 sec):** This option is incorrect because the calculated time of flight is exactly 2 seconds. A time of 2.5 seconds would require either a higher initial velocity or a different angle of projection, which is not the case.

3. **Option 4 (3 sec):** This option is incorrect because the time of flight is determined to be 2 seconds. A time of 3 seconds would suggest a different set of conditions, such as a higher initial velocity or a different angle, which do not apply to this scenario.

In conclusion, the correct answer is Option 2, as the calculations clearly show that the time of flight for the given conditions is 2 seconds.