If `H_(1) and H_(2)` be the greatest heights of a projectile in two paths for a given value of range, then the horizontal range of projectile is given by

Options

A. `(H_(1)+ H_(2))/(2)`
B. `(H_(1) + H_(2))/(4)`
C. `4sqrt(H_(1) H_(2))`
D. `4 [ H_(1) + H_(2) ] `

Answer

To find the horizontal range of a projectile given the greatest heights \( H_1 \) and \( H_2 \) for two different paths with the same horizontal range, we can follow these steps:

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

### Step 2: Express initial velocities in terms of heights
For the two paths, we can express the initial velocities in terms of the maximum heights:
\[ H_1 = \frac{u_1^2 \sin^2 \theta_1}{2g} \quad \text{and} \quad H_2 = \frac{u_2^2 \sin^2 \theta_2}{2g} \]
From these equations, we can rearrange to find:
\[ u_1 \sin \theta_1 = \sqrt{2gH_1} \quad \text{and} \quad u_2 \sin \theta_2 = \sqrt{2gH_2} \]

### Step 3: Use the range formula
The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For the two paths, we have:
\[ R_1 = \frac{u_1^2 \sin 2\theta_1}{g} \quad \text{and} \quad R_2 = \frac{u_2^2 \sin 2\theta_2}{g} \]
Since \( R_1 = R_2 \), we can set these equal to each other:
\[ \frac{u_1^2 \sin 2\theta_1}{g} = \frac{u_2^2 \sin 2\theta_2}{g} \]
This simplifies to:
\[ u_1^2 \sin 2\theta_1 = u_2^2 \sin 2\theta_2 \]

### Step 4: Relate the angles using complementary angles
Since the two angles \( \theta_1 \) and \( \theta_2 \) are complementary (i.e., \( \theta_1 + \theta_2 = 90^\circ \)), we have:
\[ \sin 2\theta_1 = 2 \sin \theta_1 \cos \theta_1 \quad \text{and} \quad \sin 2\theta_2 = 2 \sin \theta_2 \cos \theta_2 \]
Thus, we can rewrite the equation:
\[ u_1^2 \cdot 2 \sin \theta_1 \cos \theta_1 = u_2^2 \cdot 2 \sin \theta_2 \cos \theta_2 \]

### Step 5: Substitute the expressions for \( u_1 \sin \theta_1 \) and \( u_2 \sin \theta_2 \)
Substituting \( u_1 \sin \theta_1 = \sqrt{2gH_1} \) and \( u_2 \sin \theta_2 = \sqrt{2gH_2} \) into the range equation gives:
\[ \sqrt{2gH_1} \cdot 2 \cos \theta_1 = \sqrt{2gH_2} \cdot 2 \cos \theta_2 \]

### Step 6: Solve for the range
After simplifying and using the relationship between \( \tan \theta_1 \) and \( \tan \theta_2 \) (due to their complementary nature), we can derive the final expression for the range:
\[ R = 4 \sqrt{H_1 H_2} \]

### Final Answer
Thus, the horizontal range of the projectile is given by:
\[ R = 4 \sqrt{H_1 H_2} \]

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