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 step by step, we will analyze the motion of both bodies A and B using the equations of motion under gravity.

### Step 1: Identify the given data
- Mass of body A (m_A) = 4 kg
- Height from which body A is dropped (H_A) = 100 m
- Mass of body B (m_B) = 2 kg
- Height from which body B is dropped (H_B) = 50 m
- Initial velocity (u) for both bodies = 0 m/s (since they are dropped)

### Step 2: Write the equation of motion for both bodies
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) = distance fallen
- \( u \) = initial velocity
- \( a \) = acceleration (which is \( g \), the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \))
- \( t \) = time taken

For body A:
\[ H_A = 0 \cdot t_A + \frac{1}{2} g t_A^2 \]
\[ 100 = \frac{1}{2} g t_A^2 \]
\[ 100 = \frac{1}{2} (9.81) t_A^2 \]
\[ 100 = 4.905 t_A^2 \]
\[ t_A^2 = \frac{100}{4.905} \]
\[ t_A^2 \approx 20.39 \]
\[ t_A \approx \sqrt{20.39} \approx 4.51 \, \text{s} \]

For body B:
\[ H_B = 0 \cdot t_B + \frac{1}{2} g t_B^2 \]
\[ 50 = \frac{1}{2} g t_B^2 \]
\[ 50 = \frac{1}{2} (9.81) t_B^2 \]
\[ 50 = 4.905 t_B^2 \]
\[ t_B^2 = \frac{50}{4.905} \]
\[ t_B^2 \approx 10.20 \]
\[ t_B \approx \sqrt{10.20} \approx 3.19 \, \text{s} \]

### Step 3: Compare the times taken by both bodies
Now we can compare the times \( t_A \) and \( t_B \):
- \( t_A \approx 4.51 \, \text{s} \)
- \( t_B \approx 3.19 \, \text{s} \)

### Step 4: Find the ratio of times
To find the ratio of the times taken by the two bodies:
\[ \frac{t_A^2}{t_B^2} = \frac{H_A}{H_B} = \frac{100}{50} = 2 \]
This implies:
\[ t_A^2 = 2 t_B^2 \]
Taking the square root on both sides:
\[ t_A = \sqrt{2} t_B \]
Thus:
\[ t_B = \frac{t_A}{\sqrt{2}} \approx \frac{4.51}{\sqrt{2}} \approx 3.19 \, \text{s} \]

### Conclusion
Body B takes approximately 0.7 times the time required by body A to reach the ground.