A projectile is fired with speed `v_(0)` at `t=0` on a planet named ' Increasing Gravity ' . This planet is strange one, in the sense that the acceleration due to gravity increases linearly with time `t` as` g(t)=bt`, where `b` is a positive constant. 'Increase Gravity'.
If angle of projection with horizontal is `theta` then the time of flight is `:`

Options

A. `sqrt((6v_(0)sin theta)/(b))`
B. `sqrt((2v_(0)sin theta)/(b))`
C. `sqrt((3v_(0)sin theta)/(b))`
D. `sqrt((2v_(0))/(b))`

Explanation

To find the time of flight of a projectile fired on a planet where the acceleration due to gravity increases linearly with time, we can follow these steps:

### Step 1: Understand the components of the projectile motion
The projectile is fired with an initial speed \( v_0 \) at an angle \( \theta \) with the horizontal. The initial velocity can be broken down into two components:
- Horizontal component: \( v_{0x} = v_0 \cos \theta \)
- Vertical component: \( v_{0y} = v_0 \sin \theta \)

### Step 2: Define the acceleration due to gravity
The acceleration due to gravity on this planet is given by:
\[ g(t) = bt \]
where \( b \) is a positive constant and \( t \) is the time.

### Step 3: Write the equation for vertical motion
The vertical velocity \( v_y \) at any time \( t \) can be expressed as:
\[ v_y = v_{0y} - \int_0^t g(t') dt' \]
Substituting \( g(t') = bt' \):
\[ v_y = v_0 \sin \theta - \int_0^t bt' dt' \]

### Step 4: Calculate the integral
The integral \( \int_0^t bt' dt' \) evaluates to:
\[ \int_0^t bt' dt' = \frac{b t^2}{2} \]
Thus, the vertical velocity becomes:
\[ v_y = v_0 \sin \theta - \frac{b t^2}{2} \]

### Step 5: Set the vertical velocity to zero at the peak
At the peak of the projectile's motion, the vertical velocity becomes zero:
\[ 0 = v_0 \sin \theta - \frac{b t^2}{2} \]
Rearranging gives:
\[ \frac{b t^2}{2} = v_0 \sin \theta \]
Thus,
\[ t^2 = \frac{2 v_0 \sin \theta}{b} \]

### Step 6: Find the time of flight
The time of flight \( T \) is the total time for the projectile to go up and come back down. Since the time to reach the peak is \( t \), the total time of flight is:
\[ T = 2t = 2 \sqrt{\frac{2 v_0 \sin \theta}{b}} = \sqrt{\frac{8 v_0 \sin \theta}{b}} \]

### Final Answer
The time of flight \( T \) is given by:
\[ T = \sqrt{\frac{8 v_0 \sin \theta}{b}} \]