Find the ratio of maximum horizontal range and the maximum height attained by the projectile. i.e. for ` theta_(0) = 45^(@)`

Answer

To find the ratio of the maximum horizontal range to the maximum height attained by a projectile launched at an angle of \( \theta = 45^\circ \), we can follow these steps:

### Step 1: Write the formulas for range and height
The formulas for the maximum horizontal range \( R \) and maximum height \( H \) of a projectile are given by:

\[ R = \frac{u^2 \sin 2\theta}{g} \]
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

### Step 2: Substitute \( \theta = 45^\circ \)
For \( \theta = 45^\circ \):
- \( \sin 2\theta = \sin 90^\circ = 1 \)
- \( \sin \theta = \sin 45^\circ = \frac{\sqrt{2}}{2} \)

Now substituting these values into the formulas:

\[ R = \frac{u^2 \cdot 1}{g} = \frac{u^2}{g} \]

\[ H = \frac{u^2 \left(\frac{\sqrt{2}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 3: Calculate the ratio \( \frac{R}{H} \)
Now we find the ratio of the maximum horizontal range to the maximum height:

\[ \frac{R}{H} = \frac{\frac{u^2}{g}}{\frac{u^2}{4g}} = \frac{u^2}{g} \cdot \frac{4g}{u^2} = 4 \]

### Step 4: Conclusion
Thus, the ratio of the maximum horizontal range to the maximum height attained by the projectile when launched at an angle of \( 45^\circ \) is:

\[ \frac{R}{H} = 4 \]