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

Options

A. 1
B. 4
C. 1/4
D. 1/2

Explanation

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

### Step 1: Write the formulas for horizontal range and maximum height.
The formulas for the horizontal range \( R \) and maximum height \( H \) of a projectile are given by:
- Horizontal Range:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- Maximum Height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 2: Set up the ratio of horizontal range to maximum height.
We need to find the ratio \( \frac{R}{H} \):
\[ \frac{R}{H} = \frac{\frac{u^2 \sin 2\theta}{g}}{\frac{u^2 \sin^2 \theta}{2g}} \]

### Step 3: Simplify the ratio.
Cancel out \( u^2 \) and \( g \) from the numerator and denominator:
\[ \frac{R}{H} = \frac{\sin 2\theta}{\frac{1}{2} \sin^2 \theta} = \frac{2 \sin 2\theta}{\sin^2 \theta} \]

### Step 4: Substitute \( \theta = 45^\circ \).
Now, substitute \( \theta = 45^\circ \) into the equation:
- We know that \( \sin 2\theta = \sin 90^\circ = 1 \).
- Also, \( \sin 45^\circ = \frac{1}{\sqrt{2}} \).

Thus, we have:
\[ \frac{R}{H} = \frac{2 \cdot 1}{\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{2}{\frac{1}{2}} = 2 \cdot 2 = 4 \]

### Step 5: Conclusion.
The ratio of the horizontal range to the maximum height attained by the projectile when \( \theta = 45^\circ \) is:
\[ \frac{R}{H} = 4 \]