A body A of mass 4 kg is dropped from a height of 100 m. Another body B of mass 2 kg is dropped from a height of 50 m at the same time. Then :

Options

A. Both the bodies reach the ground simultaneously.
B. A takes nearly `0.7^(th)` of time required by B
C. B takes nearly `0.7^(th)` of time required by A
D. A takes double the time required by B.

Answer

To solve the problem, we need to determine the time taken by each body to reach the ground after being dropped from their respective heights. We will use the equations of motion under constant acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of body A (mA) = 4 kg (not needed for the calculation)
- Height of body A (hA) = 100 m
- Mass of body B (mB) = 2 kg (not needed for the calculation)
- Height of body B (hB) = 50 m
- Acceleration due to gravity (g) = 9.81 m/s² (approximately 10 m/s² for simplification)

2. **Use the Equation of Motion:**
The equation for the distance fallen under gravity is given by:
\[ h = ut + \frac{1}{2}gt^2 \]
Since both bodies are dropped (initial velocity \(u = 0\)), the equation simplifies to:
\[ h = \frac{1}{2}gt^2 \]

3. **Calculate Time for Body A:**
For body A:
\[ hA = \frac{1}{2}gt_A^2 \]
Substituting the values:
\[ 100 = \frac{1}{2} \cdot 9.81 \cdot t_A^2 \]
Rearranging gives:
\[ t_A^2 = \frac{200}{9.81} \]
\[ t_A = \sqrt{\frac{200}{9.81}} \approx \sqrt{20.39} \approx 4.52 \text{ seconds} \]

4. **Calculate Time for Body B:**
For body B:
\[ hB = \frac{1}{2}gt_B^2 \]
Substituting the values:
\[ 50 = \frac{1}{2} \cdot 9.81 \cdot t_B^2 \]
Rearranging gives:
\[ t_B^2 = \frac{100}{9.81} \]
\[ t_B = \sqrt{\frac{100}{9.81}} \approx \sqrt{10.19} \approx 3.19 \text{ seconds} \]

5. **Compare the Times:**
Now we can compare the times taken by both bodies:
- Time taken by body A, \(t_A \approx 4.52\) seconds
- Time taken by body B, \(t_B \approx 3.19\) seconds

6. **Conclusion:**
Body A takes longer to reach the ground than body B. The time taken by body A is greater than the time taken by body B.