For an object thrown at `45^(@)` to the horizontal, the maximum height H and horizontal range R are related as

Options

A. `R = 16H`
B. `R = 8H`
C. `R = 4H`
D. `R = 2H`

Explanation

To find the relationship between the maximum height \( H \) and the horizontal range \( R \) for an object thrown at an angle of \( 45^\circ \) to the horizontal, we can follow these steps:

### Step 1: Understand the formulas for range and height
The formulas for the 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,
- \( \theta \) is the angle of projection.

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

### Step 3: Calculate the range \( R \)
Substituting \( \theta = 45^\circ \) into the range formula:
\[ R = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} \]

### Step 4: Calculate the maximum height \( H \)
Substituting \( \theta = 45^\circ \) into the height formula:
\[ H = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 5: Relate \( R \) and \( H \)
Now, we have:
- \( R = \frac{u^2}{g} \)
- \( H = \frac{u^2}{4g} \)

To find the relationship between \( R \) and \( H \), we can express \( H \) in terms of \( R \):
\[ H = \frac{u^2}{4g} = \frac{1}{4} \cdot \frac{u^2}{g} = \frac{R}{4} \]

### Conclusion
Thus, the relationship between the maximum height \( H \) and the horizontal range \( R \) for an object thrown at \( 45^\circ \) is:
\[ R = 4H \]