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 need to find the horizontal range of a projectile given that the range is twice the maximum height attained by it. Let's break this down step by step.

### Step 1: Understand the formulas
The horizontal range \( R \) of a projectile launched with an initial speed \( u \) at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
The maximum height \( H \) attained by the projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

### Step 2: Set up the relationship
According to the problem, the horizontal range is twice the maximum height:
\[ R = 2H \]
Substituting the formulas for \( R \) and \( H \):
\[ \frac{u^2 \sin(2\theta)}{g} = 2 \left(\frac{u^2 \sin^2(\theta)}{2g}\right) \]
This simplifies to:
\[ \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{g} \]

### Step 3: Cancel out common terms
Since \( u^2 \) and \( g \) are common on both sides, we can cancel them out (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin(2\theta) = \sin^2(\theta) \]

### Step 4: Use the double angle identity
Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), we can rewrite the equation:
\[ 2 \sin(\theta) \cos(\theta) = \sin^2(\theta) \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \sin^2(\theta) - 2 \sin(\theta) \cos(\theta) = 0 \]
Factoring out \( \sin(\theta) \):
\[ \sin(\theta)(\sin(\theta) - 2\cos(\theta)) = 0 \]

### Step 6: Solve for \( \theta \)
This gives us two cases:
1. \( \sin(\theta) = 0 \) (which is not valid for projectile motion)
2. \( \sin(\theta) - 2\cos(\theta) = 0 \) or \( \sin(\theta) = 2\cos(\theta) \)

Dividing by \( \cos(\theta) \) gives:
\[ \tan(\theta) = 2 \]

### Step 7: Find \( \sin(\theta) \) and \( \cos(\theta) \)
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can represent \( \sin(\theta) \) and \( \cos(\theta) \) in terms of a right triangle:
- Let \( \sin(\theta) = 2k \) and \( \cos(\theta) = k \) for some \( k \).
- Then, \( \tan(\theta) = \frac{2k}{k} = 2 \).

Using the Pythagorean theorem:
\[ (2k)^2 + k^2 = 1 \implies 4k^2 + k^2 = 1 \implies 5k^2 = 1 \implies k^2 = \frac{1}{5} \implies k = \frac{1}{\sqrt{5}} \]
Thus:
\[ \sin(\theta) = \frac{2}{\sqrt{5}}, \quad \cos(\theta) = \frac{1}{\sqrt{5}} \]

### Step 8: Substitute back to find the range
Now substituting back into the range formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 (2 \sin(\theta) \cos(\theta))}{g} = \frac{u^2 \cdot 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}{g} \]
\[ = \frac{u^2 \cdot \frac{4}{5}}{g} = \frac{4u^2}{5g} \]

### Final Answer
Thus, the horizontal range of the projectile is:
\[ R = \frac{4u^2}{5g} \]