Prove that the maximum horizontal range is four times the maximum height attained by the projectile, when fired at an inclination so as to have maximum horizontal range.

Explanation

To prove that the maximum horizontal range \( R_{max} \) is four times the maximum height \( H_{max} \) attained by a projectile fired at an angle for maximum range, we can follow these steps:

### Step 1: Define the angle for maximum range
The angle for maximum horizontal range is \( \theta = 45^\circ \).

### Step 2: Write the formula for horizontal range
The formula for the horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Where:
- \( u \) = initial velocity
- \( g \) = acceleration due to gravity

### Step 3: Substitute \( \theta = 45^\circ \) into the range formula
Using \( \sin(90^\circ) = 1 \) (since \( 2 \times 45^\circ = 90^\circ \)):
\[ R_{max} = \frac{u^2 \cdot 1}{g} = \frac{u^2}{g} \]

### Step 4: Write the formula for maximum height
The formula for the maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

### Step 5: Substitute \( \theta = 45^\circ \) into the height formula
Using \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ H_{max} = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 6: Relate \( R_{max} \) and \( H_{max} \)
From the expressions we derived:
- \( R_{max} = \frac{u^2}{g} \)
- \( H_{max} = \frac{u^2}{4g} \)

Now, we can express \( H_{max} \) in terms of \( R_{max} \):
\[ H_{max} = \frac{1}{4} R_{max} \]

### Step 7: Prove the relationship
From the above equation, we can rearrange it to show:
\[ R_{max} = 4 H_{max} \]

Thus, we have proved that the maximum horizontal range is four times the maximum height attained by the projectile.