A football player kicks a ball at an angle of `37^(@)` to the horizontal with an initial speed of `15 ms^(-1)`. Assuming that the ball travels in a vertical plane, calculate (i) the time at which the ball reaches the highest point (ii) the maximum height reached (iii) the horizontal range of the projectile and (iv) the time for which the ball is in air.

Answer

To solve the problem step by step, we will break down each part of the question regarding the projectile motion of the football kicked by the player.

### Given Data:
- Initial speed, \( u = 15 \, \text{m/s} \)
- Angle of projection, \( \theta = 37^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (approximately)

### Step 1: Calculate the components of the initial velocity
The initial velocity can be resolved into horizontal and vertical components using trigonometric functions.

1. **Horizontal Component** (\( u_x \)):
\[ u_x = u \cdot \cos(\theta) = 15 \cdot \cos(37^\circ) \]
Using \( \cos(37^\circ) = \frac{4}{5} \):
\[ u_x = 15 \cdot \frac{4}{5} = 12 \, \text{m/s} \]

2. **Vertical Component** (\( u_y \)):
\[ u_y = u \cdot \sin(\theta) = 15 \cdot \sin(37^\circ) \]
Using \( \sin(37^\circ) = \frac{3}{5} \):
\[ u_y = 15 \cdot \frac{3}{5} = 9 \, \text{m/s} \]

### Step 2: Calculate the time to reach the highest point
At the highest point, the vertical component of the velocity becomes zero. We can use the first equation of motion:
\[ v_y = u_y - g \cdot t \]
Setting \( v_y = 0 \):
\[ 0 = 9 - 10 \cdot t \]
Solving for \( t \):
\[ t = \frac{9}{10} = 0.9 \, \text{s} \]

### Step 3: Calculate the maximum height reached
We can use the second equation of motion to find the maximum height (\( H \)):
\[ H = u_y \cdot t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ H = 9 \cdot 0.9 - \frac{1}{2} \cdot 10 \cdot (0.9)^2 \]
Calculating:
\[ H = 8.1 - 4.05 = 4.05 \, \text{m} \]

### Step 4: Calculate the total time of flight
The total time of flight (\( T \)) is twice the time to reach the highest point:
\[ T = 2 \cdot t = 2 \cdot 0.9 = 1.8 \, \text{s} \]

### Step 5: Calculate the horizontal range
The horizontal range (\( R \)) can be calculated using:
\[ R = u_x \cdot T \]
Substituting the values:
\[ R = 12 \cdot 1.8 = 21.6 \, \text{m} \]

### Summary of Results:
1. Time to reach the highest point: \( 0.9 \, \text{s} \)
2. Maximum height reached: \( 4.05 \, \text{m} \)
3. Horizontal range of the projectile: \( 21.6 \, \text{m} \)
4. Total time of flight: \( 1.8 \, \text{s} \)