A force of magnitude 2mg(1-ay) starts acting in the vertically upward direction on a body of mass m placed on earth's surface where y is the height of the object above the ground during ascent and 'a' is a positive constant. Total height through which the body ascends is

Options

A. `1//a`
B. `1//2a`
C. `2//3a`
D. `2//a`

Answer

To solve the problem, we need to analyze the forces acting on the body and derive the height through which the body ascends.

### Step-by-Step Solution:

1. **Identify the Forces**:
The force acting on the body is given as \( F = 2mg(1 - ay) \). The weight of the body acting downward is \( mg \). Therefore, the net force \( F_{net} \) acting on the body can be expressed as:
\[ F_{net} = F - mg = 2mg(1 - ay) - mg \]

2. **Simplify the Net Force**:
Simplifying the expression for the net force:
\[ F_{net} = 2mg(1 - ay) - mg = mg(2(1 - ay) - 1) = mg(2 - 2ay - 1) = mg(1 - 2ay) \]

3. **Apply Newton's Second Law**:
According to Newton's second law, the net force is also equal to the mass times acceleration (\( F_{net} = ma \)). Thus:
\[ mg(1 - 2ay) = ma \]
Dividing both sides by \( m \) gives:
\[ a = g(1 - 2ay) \]

4. **Relate Acceleration to Velocity**:
We can express acceleration \( a \) in terms of velocity \( v \) and height \( y \) using the relationship:
\[ a = v \frac{dv}{dy} \]
Therefore, we have:
\[ v \frac{dv}{dy} = g(1 - 2ay) \]

5. **Integrate to Find Velocity**:
Rearranging gives:
\[ v dv = g(1 - 2ay) dy \]
Now, we integrate both sides. The left side integrates from \( 0 \) to \( v \) and the right side from \( 0 \) to \( y \):
\[ \int_0^v v dv = g \int_0^y (1 - 2ay) dy \]
The left side becomes:
\[ \frac{v^2}{2} \bigg|_0^v = \frac{v^2}{2} \]
The right side becomes:
\[ g \left( y - a y^2 \right) \bigg|_0^y = g \left( y - a y^2 \right) \]

6. **Set the Integrals Equal**:
Setting the two integrals equal gives:
\[ \frac{v^2}{2} = g \left( y - a y^2 \right) \]

7. **Find the Maximum Height**:
At the maximum height, the velocity \( v = 0 \). Therefore, we set the left side to zero:
\[ 0 = g \left( y - a y^2 \right) \]
This implies:
\[ y(1 - ay) = 0 \]
Thus, \( y = 0 \) or \( y = \frac{1}{a} \).

8. **Conclusion**:
The total height through which the body ascends is:
\[ y = \frac{1}{a} \]