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

Options

A. Continues to fall down
B. falls down retardation & goes up again with acceleration
C. falls down with uniform velocity
D. raises up with acceleration

Answer

To solve the problem step by step, we will analyze the motion of the body falling freely from a height 'h' and then consider the effects of reversing the acceleration due to gravity.

### Step 1: Analyze the free fall motion
When the body is dropped from height 'h', it falls under the influence of gravity. The initial velocity (u) is 0 m/s, and the acceleration (g) is acting downwards.

Using the equation of motion:
\[ s = ut + \frac{1}{2} g t^2 \]
where:
- \( s \) = distance fallen (which will be -h, as we consider downward direction negative),
- \( u \) = initial velocity = 0,
- \( g \) = acceleration due to gravity,
- \( t \) = time = 2 seconds.

Substituting the values:
\[ -h = 0 + \frac{1}{2} g (2^2) \]
\[ -h = 2g \]
Thus, we find:
\[ h = -2g \]

### Step 2: Calculate the velocity after 2 seconds
The velocity of the body after 2 seconds can be calculated using the equation:
\[ v = u + gt \]
Substituting the values:
\[ v = 0 + g(2) = 2g \]
Thus, the velocity of the body after 2 seconds is \( 2g \) downwards.

### Step 3: Reverse the acceleration due to gravity
Now, we reverse the direction of gravity. This means that the acceleration due to gravity (g) will now act upwards.

### Step 4: Analyze the motion after reversing gravity
At the moment gravity is reversed, the body has a downward velocity of \( 2g \). Since the acceleration due to gravity is now acting upwards, it will act as a retardation on the body.

Using the equation of motion again:
\[ v = u + at \]
Here, \( v \) will become 0 when the body stops moving downwards, \( u = -2g \) (downward), and \( a = g \) (upward).

Setting \( v = 0 \):
\[ 0 = -2g + g t \]
Rearranging gives:
\[ gt = 2g \implies t = 2 \text{ seconds} \]

### Step 5: Determine the position after 2 seconds of upward motion
After 2 seconds, the body will have come to rest. The distance traveled upwards during this time can be calculated as:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ s = (-2g)(2) + \frac{1}{2}(g)(2^2) \]
\[ s = -4g + 2g = -2g \]

### Conclusion
After reversing the acceleration due to gravity, the body first decelerates to a stop and then begins to move upwards. Thus, the body falls down with retardation and then goes up with acceleration.

### Final Answer
The body falls down with retardation and then goes up with acceleration.