A body falls freely from a height 'h' after two seconds if accelaration due to gravity is reversed the body

Options

A. continues to fall down
B. falls down with retardation & goes up again with acceleration after some time
C. falls down with uniform velocity
D. raises up with acceleration

Answer

To solve the problem of a body falling freely from a height 'h' and then experiencing a reversal of acceleration due to gravity after two seconds, we will analyze the motion step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- A body is falling freely from a height 'h'.
- The acceleration due to gravity (g) acts downwards.
- The time of fall before the acceleration is reversed is given as 2 seconds.

2. **Calculating the Velocity After 2 Seconds**:
- The formula for the velocity of a freely falling body after time \( t \) is:
\[ v = u + gt \]
- Here, the initial velocity \( u = 0 \) (since it starts from rest), \( g \) is the acceleration due to gravity, and \( t = 2 \) seconds.
- Therefore, the velocity after 2 seconds is:
\[ v = 0 + g \cdot 2 = 2g \]

3. **Position of the Body After 2 Seconds**:
- The distance fallen in 2 seconds can be calculated using the formula:
\[ s = ut + \frac{1}{2}gt^2 \]
- Substituting \( u = 0 \), \( g \), and \( t = 2 \):
\[ s = 0 + \frac{1}{2}g(2^2) = 2g \]
- Thus, after 2 seconds, the body has fallen a distance of \( 2g \).

4. **Height Above Ground After 2 Seconds**:
- If the initial height is \( h \), the height of the body above the ground after 2 seconds is:
\[ h - 2g \]

5. **Reversal of Acceleration**:
- After 2 seconds, the acceleration due to gravity is reversed (i.e., it acts upwards).
- The body now has an initial velocity of \( 2g \) upwards and an upward acceleration of \( -g \).

6. **Motion After Reversal**:
- The body will continue to move upward until its velocity becomes zero.
- The time taken to reach the maximum height can be calculated using:
\[ v = u + at \]
- Setting \( v = 0 \) (at the maximum height), \( u = 2g \), and \( a = -g \):
\[ 0 = 2g - gt \implies gt = 2g \implies t = 2 \text{ seconds} \]
- Therefore, it takes another 2 seconds to reach the maximum height.

7. **Conclusion**:
- The body first falls down for 2 seconds, then it goes up for another 2 seconds with retardation due to the reversed acceleration. Therefore, the final motion is:
- Falls down with retardation and goes up with acceleration after some time.

### Final Answer:
The body falls down with retardation and goes up with acceleration after some time, which corresponds to option (b).