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(2H sin theta))/g`
D. `(sqrt(2H sin theta))/g`

Explanation

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

### Step 1: Understand the Motion
The stone is thrown at an angle \( \theta \) and follows a projectile motion path. The time of flight can be divided into two parts: the time taken to reach the maximum height and the time taken to descend back to the ground.

### Step 2: Analyze the Vertical Motion
At the maximum height \( H \), the vertical component of the velocity becomes zero. We can use the kinematic equation for vertical motion:
\[ H = U_y \cdot t - \frac{1}{2} g t^2 \]
where:
- \( H \) is the maximum height,
- \( U_y \) is the initial vertical velocity,
- \( t \) is the time taken to reach the maximum height,
- \( g \) is the acceleration due to gravity.

### Step 3: Determine Initial Vertical Velocity
The initial vertical velocity \( U_y \) can be expressed as:
\[ U_y = U \sin \theta \]
where \( U \) is the initial velocity of the stone.

### Step 4: Time to Reach Maximum Height
At maximum height, the vertical velocity is zero. Using the equation:
\[ 0 = U_y - g t \]
we can solve for \( t \):
\[ t = \frac{U_y}{g} = \frac{U \sin \theta}{g} \]
This \( t \) is the time taken to reach maximum height, which is \( \frac{T}{2} \) (where \( T \) is the total time of flight).

### Step 5: Relate Height to Time
Substituting \( U_y \) into the height equation:
\[ H = U \sin \theta \cdot \left(\frac{U \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{U \sin \theta}{g}\right)^2 \]
This simplifies to:
\[ H = \frac{(U \sin \theta)^2}{2g} \]

### Step 6: Solve for Total Time of Flight
From the height equation, we can express \( U \sin \theta \) in terms of \( H \):
\[ (U \sin \theta)^2 = 2gH \]
Now, substituting back to find \( T \):
\[ T = 2t = 2 \cdot \frac{U \sin \theta}{g} \]
Using \( U \sin \theta = \sqrt{2gH} \):
\[ T = 2 \cdot \frac{\sqrt{2gH}}{g} = 2\sqrt{\frac{2H}{g}} \]

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