A body is dropped from a height of 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity ( Take `g=10 m//s^(2)` )

Options

A. 19.6 m/s
B. 20 m/s
C. 1.96 m/s
D. 196 m/s

Answer

To solve the problem step by step, we will analyze the motion of the body dropped from a height of 39.2 meters, considering the conditions given in the question.

### Step 1: Determine the Half Distance
The total height from which the body is dropped is 39.2 m. We first find half of this height:
\[ \text{Half Distance} = \frac{39.2}{2} = 19.6 \text{ m} \]

### Step 2: Use the Kinematic Equation
When the body falls the first half of the distance (19.6 m), it is under the influence of gravity. We can use the kinematic equation to find the final velocity just before it reaches the half distance:
\[ v^2 = u^2 + 2as \]
where:
- \(v\) = final velocity
- \(u\) = initial velocity (0 m/s, since the body is dropped)
- \(a\) = acceleration due to gravity (10 m/s²)
- \(s\) = distance fallen (19.6 m)

### Step 3: Substitute Values into the Equation
Substituting the known values into the equation:
\[ v^2 = 0^2 + 2 \cdot 10 \cdot 19.6 \]
\[ v^2 = 0 + 392 \]
\[ v^2 = 392 \]

### Step 4: Calculate the Final Velocity
Now, we take the square root to find the final velocity \(v\):
\[ v = \sqrt{392} \]
Calculating the square root:
\[ v = \sqrt{392} \approx 19.8 \text{ m/s} \]

### Step 5: Analyze the Motion After Half Distance
After the body crosses the half distance (19.6 m), the problem states that the acceleration due to gravity ceases to act. Therefore, the body will continue to fall with a constant velocity of approximately 19.8 m/s until it hits the ground.

### Conclusion
The body will hit the ground with a velocity of approximately **19.8 m/s**.

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