A stone is thrown at an angle `theta` to the horizontal reaches a maximum height H. Then the time of flight of stone will be:

Options

A. `sqrt((2H)/(g))`
B. `2sqrt((2H)/(g))`
C. `2sqrt((2Hsintheta)/(g))`
D. `sqrt((2Hsintheta)/(g))`

Explanation

To find the time of flight of a stone thrown at an angle \(\theta\) to the horizontal, we can follow these steps:

### Step 1: Understand the problem
The stone is thrown at an angle \(\theta\) and reaches a maximum height \(H\). We need to determine the total time of flight for the stone.

### Step 2: Use the formula for maximum height
The maximum height \(H\) reached by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \(u\) is the initial velocity and \(g\) is the acceleration due to gravity.

### Step 3: Rearranging the formula for initial velocity component
From the maximum height formula, we can express \(u \sin \theta\):
\[ u \sin \theta = \sqrt{2gH} \]

### Step 4: Use the formula for time of flight
The total time of flight \(T\) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]

### Step 5: Substitute the expression for \(u \sin \theta\)
Now, substitute \(u \sin \theta\) from Step 3 into the time of flight formula:
\[ T = \frac{2 \sqrt{2gH}}{g} \]

### Step 6: Simplify the expression
Now simplify the expression:
\[ T = \frac{2\sqrt{2H}}{\sqrt{g}} \]

### Final Result
Thus, the time of flight \(T\) of the stone is:
\[ T = 2\sqrt{\frac{2H}{g}} \]