A player in the football team tries to kick a football so that it stays in the air for a long "hang time". If the ball is kicked with an initial velocity of 25.0 m/s at an angle of `60.0^(@)` above the ground, what is the "hang time"?

Options

A. 4.42 s
B. 3.36 s
C. 2.21 s
D. 1.68 s

Answer

To solve the problem of finding the "hang time" or time of flight of a football kicked with an initial velocity of 25.0 m/s at an angle of 60.0 degrees above the ground, we can use the formula for the time of flight in projectile motion.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Initial velocity (u) = 25.0 m/s
- Angle of projection (θ) = 60.0 degrees
- Acceleration due to gravity (g) = 9.8 m/s² (we can also use 10 m/s² for simplicity in calculations)

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

3. **Calculate \(\sin \theta\)**:
For θ = 60 degrees, we know:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \]

4. **Substitute the Values into the Formula**:
Using the values we have:
\[ T = \frac{2 \times 25.0 \, \text{m/s} \times \sin 60^\circ}{g} \]
Substituting \(\sin 60^\circ\):
\[ T = \frac{2 \times 25.0 \, \text{m/s} \times 0.866}{9.8 \, \text{m/s}^2} \]

5. **Calculate the Numerator**:
\[ 2 \times 25.0 \times 0.866 = 43.3 \, \text{m/s} \]

6. **Divide by g**:
Using \(g = 9.8 \, \text{m/s}^2\):
\[ T = \frac{43.3}{9.8} \approx 4.41 \, \text{s} \]

7. **Final Result**:
The hang time or time of flight of the football is approximately **4.41 seconds**.