A football player throws a ball with a velocity of `50 m//s` at an angle 30 degrees from the horizontal. The ball remains in the air for `(g = 10m//s^(2))`

Options

A. `2.5 s`
B. `1.25s`
C. `5s`
D. `0.625s`

Explanation

To find the time a football remains in the air when thrown with a velocity of 50 m/s at an angle of 30 degrees, we can use the formula for the time of flight of a projectile:

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity, \( u = 50 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

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

3. **Calculate \( \sin 30^\circ \):**
We know that:
\[ \sin 30^\circ = \frac{1}{2} \]

4. **Substitute the values into the formula:**
Plugging in the values into the time of flight formula:
\[ T = \frac{2 \times 50 \times \sin 30^\circ}{10} \]
\[ T = \frac{2 \times 50 \times \frac{1}{2}}{10} \]

5. **Simplify the equation:**
\[ T = \frac{50}{10} = 5 \, \text{seconds} \]

6. **Conclusion:**
The time the ball remains in the air is \( T = 5 \, \text{seconds} \).

### Final Answer:
The ball remains in the air for **5 seconds**.

---