An object is thrown along a direction inclined at an angle of `45^(@)` with the horizontal direction. The horizontal range of the particle is equal to

Options

A. vertical height
B. twice the vertical height
C. thrice the vertice height
D. four times the vertical height

Explanation

To find the horizontal range of an object thrown at an angle of 45 degrees with the horizontal, we can use the following steps:

### Step 1: Understand the formulas for range and height
The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

The maximum height \( H \) attained by the projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

### Step 2: Substitute the angle into the formulas
Since the angle \( \theta \) is given as \( 45^\circ \):
- For the range:
\[ R = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} \]
(since \( \sin(90^\circ) = 1 \))

- For the height:
\[ H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 3: Relate the range to the height
Now we have:
- \( R = \frac{u^2}{g} \)
- \( H = \frac{u^2}{4g} \)

To find the relationship between \( R \) and \( H \), we can express \( R \) in terms of \( H \):
\[ H = \frac{u^2}{4g} \implies u^2 = 4gH \]
Substituting this into the equation for \( R \):
\[ R = \frac{4gH}{g} = 4H \]

### Conclusion
Thus, the horizontal range \( R \) is equal to four times the vertical height \( H \):
\[ R = 4H \]

### Final Answer
The horizontal range of the particle is equal to **four times the vertical height**.

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