A body is thrown with a velocity of 9.8 m/s making an angle of 30° with the horizontal. It will hit the ground after a time

The correct option/answer is 1 s.

Let us solve the problem step-by-step using the principles of projectile motion.

1. Identify the given values:
- Initial velocity of projection ($u$) = 9.8 m/s
- Angle of projection with the horizontal ($\theta$) = 30°
- Acceleration due to gravity ($g$) is approximately 9.8 m/s²

2. Understand the concept:
The total time the body remains in the air before hitting the ground is called the time of flight ($T$).
For a projectile launched from ground level and returning to the same ground level, the formula for the time of flight is:

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

3. Calculation:
Substitute the given values into the formula:

$T=\frac{2\times 9.8\times \sin (30°)}{9.8}$

We know that $\sin (30°)=0.5$ (or $\frac{1}{2}$).
We can cancel the 9.8 in the numerator and the denominator:

$T=2\times 0.5$

$T=1\text{s}$

Thus, the body will hit the ground after 1 s.