A player kicks a ball at an angle of `37^(@)` to the horizontal with an initial speed of ` 15 ms^(-1)` . Find its time of fight.

Answer

To find the time of flight of the ball kicked at an angle of \(37^\circ\) with an initial speed of \(15 \, \text{m/s}\), we can use the formula for the time of flight in projectile motion:

\[ T = \frac{2u \sin \theta}{g} \]

Where:
- \(T\) is the time of flight,
- \(u\) is the initial speed,
- \(\theta\) is the angle of projection,
- \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)).

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial speed, \(u = 15 \, \text{m/s}\)
- Angle of projection, \(\theta = 37^\circ\)
- Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)

2. **Calculate \(\sin \theta\):**
- We need to find \(\sin 37^\circ\). From trigonometric tables or a calculator, we know:
\[ \sin 37^\circ \approx 0.6 \]

3. **Substitute the values into the formula:**
\[ T = \frac{2 \times 15 \times \sin 37^\circ}{10} \]
\[ T = \frac{2 \times 15 \times 0.6}{10} \]

4. **Calculate the numerator:**
\[ 2 \times 15 \times 0.6 = 18 \]

5. **Divide by \(g\):**
\[ T = \frac{18}{10} = 1.8 \, \text{seconds} \]

### Final Answer:
The time of flight of the ball is \(1.8 \, \text{seconds}\).

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