For an object projected from ground with speed u horizontal range is two times the maximum height attained by it. The horizontal range of object is

Options

A. `(2u^(2))/(3g)`
B. `(3u^(2))/(4g)`
C. `(3u^(2))/(2g)`
D. `(4u^(2))/(5g)`

Answer

To solve the problem, we will use the equations of motion for projectile motion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given that the horizontal range (R) of a projectile is twice the maximum height (H) attained by it. We need to find the horizontal range (R) in terms of the initial speed (u).

2. **Formulas for Maximum Height (H) and Range (R)**:
- The maximum height (H) attained by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- The horizontal range (R) of the projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

3. **Setting Up the Relationship**:
According to the problem, we have:
\[ R = 2H \]
Substituting the formulas for R and H into this equation gives:
\[ \frac{u^2 \sin 2\theta}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right) \]

4. **Simplifying the Equation**:
Simplifying the right side:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{g} \]
We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

5. **Using the Double Angle Identity**:
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Therefore, we can rewrite the equation as:
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \sin^2 \theta - 2 \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 2 \cos \theta) = 0 \]

7. **Finding the Angles**:
This gives us two cases:
- Case 1: \( \sin \theta = 0 \) (which is not valid for projectile motion)
- Case 2: \( \sin \theta = 2 \cos \theta \)
Dividing both sides by \( \cos \theta \) gives:
\[ \tan \theta = 2 \]

8. **Calculating R**:
Now, we can find \( R \) using \( \tan \theta = 2 \):
- From \( \tan \theta = \frac{2}{1} \), we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}} \]
- Now substituting \( \sin \theta \) into the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 \cdot 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}{g} = \frac{u^2 \cdot \frac{4}{5}}{g} \]
- Thus, the horizontal range \( R \) is:
\[ R = \frac{4u^2}{5g} \]

### Final Answer:
The horizontal range of the object is \( R = \frac{4u^2}{5g} \).