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

Options

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

Answer

To solve the problem, we need to find the horizontal range of an object thrown at an angle of 45 degrees with the horizontal. We will use the formulas for vertical height and horizontal range in projectile motion.

### Step-by-Step Solution:

1. **Identify the angle of projection**:
The angle of projection, θ, is given as 45 degrees.

2. **Use the formula for vertical height (H)**:
The vertical height \( H \) achieved by the projectile can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

3. **Substitute θ = 45 degrees into the height formula**:
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), we have:
\[ H = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

4. **Use the formula for horizontal range (R)**:
The horizontal range \( R \) can be calculated using the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For \( \theta = 45^\circ \), we have \( \sin 90^\circ = 1 \):
\[ R = \frac{u^2 \cdot 1}{g} = \frac{u^2}{g} \]

5. **Relate R to H**:
We can express \( R \) in terms of \( H \):
\[ R = \frac{u^2}{g} = \frac{u^2}{4g} \cdot 4 = 4H \]
Therefore, the horizontal range \( R \) is equal to 4 times the vertical height \( H \).

6. **Conclusion**:
From the calculations, we find that:
\[ R = 4H \]
Thus, the horizontal range of the object is **four times the vertical height**.

### Final Answer:
The horizontal range of the object is **4 times the vertical height**.