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 maximum height attained is

Options

A. `(1)/(3) ((v_(0)sintheta)^(3//2))/(sqrt(b))`
B. `(4)/(3) ((v_(0)sintheta)^(3//2))/(sqrt(b))`
C. `((2v_(0)sintheta)^(3//2))/(3sqrt(b))`
D. None of these

Explanation

To find the maximum height attained by a projectile fired on a planet with increasing gravity, we can follow these steps:

### Step 1: Understand the motion equations
The acceleration due to gravity is given as \( g(t) = bt \), where \( b \) is a positive constant. The vertical motion of the projectile can be described by the equations of motion under variable acceleration.

### Step 2: Set up the vertical motion equation
The vertical component of the initial velocity is given by:
\[ v_{0y} = v_0 \sin \theta \]
The vertical position \( y \) as a function of time \( t \) can be expressed as:
\[ y(t) = v_{0y} t - \int_0^t g(t') dt' \]
Substituting \( g(t) = bt' \):
\[ y(t) = v_0 \sin \theta \cdot t - \int_0^t bt' dt' = v_0 \sin \theta \cdot t - \frac{b t^2}{2} \]

### Step 3: Find the time of maximum height
To find the maximum height, we need to determine when the vertical velocity becomes zero. The vertical velocity \( v_y(t) \) is given by:
\[ v_y(t) = v_{0y} - \int_0^t g(t') dt' = v_0 \sin \theta - bt \]
Setting \( v_y(t) = 0 \) for maximum height:
\[ 0 = v_0 \sin \theta - bt \implies t = \frac{v_0 \sin \theta}{b} \]

### Step 4: Substitute time back into the height equation
Now, substitute \( t = \frac{v_0 \sin \theta}{b} \) into the height equation:
\[ y\left(\frac{v_0 \sin \theta}{b}\right) = v_0 \sin \theta \cdot \frac{v_0 \sin \theta}{b} - \frac{b}{2} \left(\frac{v_0 \sin \theta}{b}\right)^2 \]
Calculating each term:
1. The first term:
\[ v_0 \sin \theta \cdot \frac{v_0 \sin \theta}{b} = \frac{v_0^2 \sin^2 \theta}{b} \]
2. The second term:
\[ -\frac{b}{2} \cdot \frac{(v_0 \sin \theta)^2}{b^2} = -\frac{v_0^2 \sin^2 \theta}{2b} \]

### Step 5: Combine the terms
Now, combine the two terms:
\[ y_{max} = \frac{v_0^2 \sin^2 \theta}{b} - \frac{v_0^2 \sin^2 \theta}{2b} = \frac{v_0^2 \sin^2 \theta}{2b} \]

### Step 6: Final expression for maximum height
Thus, the maximum height attained by the projectile is:
\[ y_{max} = \frac{v_0^2 \sin^2 \theta}{2b} \]