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

Options

A. 3 s
B. 2s
C. 1.5 s
D. 1 s

Explanation

To solve the problem of how long a body thrown with a velocity of 9.8 m/s at an angle of 30 degrees will remain in the air, we can use the formula for the total time of flight in projectile motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity (u) = 9.8 m/s
- Angle of projection (θ) = 30 degrees
- Acceleration due to gravity (g) = 9.8 m/s² (approximately)

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

3. **Calculate the Sine of the Angle:**
- For θ = 30 degrees, we know:
\[ \sin 30^\circ = \frac{1}{2} \]

4. **Substitute the Values into the Formula:**
Now, substituting the values into the formula:
\[ T = \frac{2 \times 9.8 \times \sin 30^\circ}{9.8} \]
\[ T = \frac{2 \times 9.8 \times \frac{1}{2}}{9.8} \]

5. **Simplify the Expression:**
The 2 in the numerator and the \(\frac{1}{2}\) will cancel out:
\[ T = \frac{9.8}{9.8} = 1 \text{ second} \]

6. **Final Answer:**
The total time of motion (time the body will remain in the air) is:
\[ T = 1 \text{ second} \]