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 solve the problem of finding the time of flight of a stone thrown at an angle \( \theta \) to the horizontal that reaches a maximum height \( H \), we can follow these steps:

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

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

### Step 3: Square the time of flight equation
Squaring the time of flight equation gives:
\[ T^2 = \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{4u^2 \sin^2 \theta}{g^2} \]

### Step 4: Relate time of flight and maximum height
Now, we can relate the two equations. From the maximum height equation, we can express \( u^2 \sin^2 \theta \):
\[ u^2 \sin^2 \theta = 2gH \]

### Step 5: Substitute into the time of flight equation
Substituting \( u^2 \sin^2 \theta \) into the squared time of flight equation:
\[ T^2 = \frac{4(2gH)}{g^2} = \frac{8gH}{g^2} = \frac{8H}{g} \]

### Step 6: Solve for time of flight \( T \)
Taking the square root of both sides gives:
\[ T = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}} \]

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