A body falls from a certain bhight. Two seconds later another body falls from the same height. How long after the beginning of motion of the first body is the distance between the bodies twice the distance at the moment the second body starts to fall ?

Options

A. 3 s
B. 10 s
C. 15 s
D. 20 s

Answer

To solve the problem step by step, we will analyze the motion of both bodies and set up the equations based on the information given.

### Step 1: Understand the motion of the first body
The first body falls from a height \( h \) with an initial velocity \( u_1 = 0 \). The distance it falls after \( t_1 \) seconds is given by the equation of motion:
\[ h_1 = \frac{1}{2} g t_1^2 \]
where \( g \) is the acceleration due to gravity.

### Step 2: Determine the time for the first body
Since the second body starts falling 2 seconds after the first body, we can denote the time when the first body has fallen as \( T \). Therefore, the time for the first body is:
\[ t_1 = T \]
Thus, the distance fallen by the first body becomes:
\[ h_1 = \frac{1}{2} g T^2 \]

### Step 3: Motion of the second body
The second body starts falling after 2 seconds, so its time of fall \( t_2 \) is:
\[ t_2 = T - 2 \]
The distance fallen by the second body is:
\[ h_2 = \frac{1}{2} g (T - 2)^2 \]

### Step 4: Set up the distance condition
We need to find \( T \) such that the distance between the two bodies is twice the distance fallen by the second body at the moment it starts falling. The distance between the two bodies at time \( T \) is:
\[ \text{Distance between bodies} = h_1 - h_2 \]
We need this distance to be twice the distance fallen by the second body at \( t = 2 \) seconds:
\[ h_1 - h_2 = 2 \cdot h_2 \text{ (at the moment the second body starts falling)} \]

### Step 5: Calculate the distances
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{1}{2} g T^2 - \frac{1}{2} g (T - 2)^2 = 2 \cdot \frac{1}{2} g (T - 2)^2 \]
This simplifies to:
\[ \frac{1}{2} g T^2 - \frac{1}{2} g (T^2 - 4T + 4) = g (T - 2)^2 \]
\[ \frac{1}{2} g T^2 - \frac{1}{2} g T^2 + 2gT - 2g = g (T^2 - 4T + 4) \]
\[ 2gT - 2g = g (T^2 - 4T + 4) \]

### Step 6: Simplify the equation
Dividing through by \( g \) (assuming \( g \neq 0 \)):
\[ 2T - 2 = T^2 - 4T + 4 \]
Rearranging gives:
\[ T^2 - 6T + 6 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ T = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3} \]
Since time cannot be negative, we take:
\[ T = 3 + \sqrt{3} \text{ (approximately 4.73 seconds)} \]

### Step 8: Conclusion
The time after the beginning of motion of the first body when the distance between the bodies is twice the distance at the moment the second body starts to fall is approximately \( 4.73 \) seconds.