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'
At what angle with horizontal should the projectivle be fired so that it travels the maximum horizontal distance `:`

Options

A. `theta=tan^(-1) (1)/(2)`
B. `theta=tan^(-1) (1)/(sqrt(2))`
C. `theta=tan^(-1)sqrt(2)`
D. `theta=tan^(-1)2`

Explanation

To solve the problem of determining the angle at which a projectile should be fired to achieve maximum horizontal distance on a planet with increasing gravity, we can follow these steps:

### Step 1: Understand the Problem
The projectile is fired with an initial speed \( v_0 \) at an angle \( \theta \) with respect to the horizontal. The acceleration due to gravity is not constant; it increases linearly with time as \( g(t) = bt \), where \( b \) is a positive constant.

### Step 2: Break Down the Initial Velocity
The initial velocity \( v_0 \) can be broken down into horizontal and vertical components:
- Horizontal component: \( v_{0x} = v_0 \cos \theta \)
- Vertical component: \( v_{0y} = v_0 \sin \theta \)

### Step 3: Analyze Vertical Motion
The vertical motion is affected by the increasing gravity. The vertical velocity \( v_y \) at any time \( t \) can be expressed as:
\[ v_y = v_0 \sin \theta - \int_0^t g(t') dt' = v_0 \sin \theta - \int_0^t bt' dt' = v_0 \sin \theta - \frac{1}{2} b t^2 \]

### Step 4: Determine the Time of Flight
The projectile will reach its maximum height when \( v_y = 0 \):
\[ 0 = v_0 \sin \theta - \frac{1}{2} b t^2 \]
Solving for \( t \) gives:
\[ t = \sqrt{\frac{2 v_0 \sin \theta}{b}} \]
This is the time to reach the maximum height.

### Step 5: Calculate Total Time of Flight
Since the projectile will take the same amount of time to descend back to the original height, the total time of flight \( T \) is:
\[ T = 2 \sqrt{\frac{2 v_0 \sin \theta}{b}} = \sqrt{\frac{8 v_0 \sin \theta}{b}} \]

### Step 6: Calculate Horizontal Range
The horizontal range \( R \) can be calculated as:
\[ R = v_{0x} \cdot T = (v_0 \cos \theta) \cdot \sqrt{\frac{8 v_0 \sin \theta}{b}} \]
\[ R = v_0 \cos \theta \cdot \sqrt{\frac{8 v_0 \sin \theta}{b}} \]
\[ R = \frac{v_0 \sqrt{8 v_0}}{\sqrt{b}} \cdot \cos \theta \cdot \sqrt{\sin \theta} \]

### Step 7: Maximize the Range
To find the angle \( \theta \) that maximizes the range \( R \), we differentiate \( R \) with respect to \( \theta \) and set the derivative to zero:
\[ \frac{dR}{d\theta} = 0 \]
Using the product and chain rule, we can derive the condition for maximum range.

### Step 8: Solve the Derivative
After differentiating and simplifying, we arrive at the condition:
\[ \frac{\cos^2 \theta}{2 \sqrt{\sin \theta}} = \frac{3}{2} \sin^{3/2} \theta \]
This leads to:
\[ \tan^2 \theta = \frac{1}{2} \]
Thus,
\[ \tan \theta = \frac{1}{\sqrt{2}} \]
Therefore, the angle \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \]

### Final Answer
The angle at which the projectile should be fired to achieve maximum horizontal distance is:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \]