A particle is dropped from point `A` at a certain height from ground. It falls freely and passes through thre points `B, C` and `D` with `BC=CD`. The time taken by the particle to move from `B` to `CD` is `2s` and from `C` to `D` is `1s`. The time taken to move from `A` to `B` is s

Options

A. `0.5s`
B. `1.5s`
C. `0.75s`
D. `0.25s`

Answer

To solve the problem step by step, we will use the equations of motion for a freely falling body.

### Step 1: Understand the motion from A to B
The particle is dropped from point A, so the initial velocity (u) at point A is 0. The distance from A to B is denoted as \( h_1 \).

Using the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
For the motion from A to B:
\[ h_1 = 0 \cdot t + \frac{1}{2}g t^2 \quad \text{(where } t \text{ is the time taken from A to B)} \]
This simplifies to:
\[ h_1 = \frac{1}{2}gt^2 \quad \text{(1)} \]

### Step 2: Analyze the motion from B to C
The time taken to move from B to C is given as 2 seconds. The distance BC is denoted as \( h_2 \).

Using the second equation of motion again:
\[ h_2 = v_B \cdot t + \frac{1}{2}gt^2 \]
Where \( v_B \) is the velocity at point B. We can find \( v_B \) using the third equation of motion:
\[ v_B^2 = u^2 + 2as \quad \Rightarrow \quad v_B^2 = 0 + 2gh_1 \quad \Rightarrow \quad v_B = \sqrt{2gh_1} \quad \text{(2)} \]
Now substituting \( v_B \) into the equation for \( h_2 \):
\[ h_2 = \sqrt{2gh_1} \cdot 2 + \frac{1}{2}g(2^2) \]
This simplifies to:
\[ h_2 = 2\sqrt{2gh_1} + 2g \quad \text{(3)} \]

### Step 3: Analyze the motion from C to D
The time taken to move from C to D is given as 1 second. The distance CD is also \( h_2 \).

Using the second equation of motion:
\[ h_2 = v_C \cdot t + \frac{1}{2}gt^2 \]
Where \( v_C \) is the velocity at point C. We can find \( v_C \) using:
\[ v_C = v_B + gt_B \quad \Rightarrow \quad v_C = \sqrt{2gh_1} + g \cdot 2 \quad \Rightarrow \quad v_C = \sqrt{2gh_1} + 2g \quad \text{(4)} \]
Now substituting \( v_C \) into the equation for \( h_2 \):
\[ h_2 = \left(\sqrt{2gh_1} + 2g\right) \cdot 1 + \frac{1}{2}g(1^2) \]
This simplifies to:
\[ h_2 = \sqrt{2gh_1} + 2g + \frac{1}{2}g \quad \Rightarrow \quad h_2 = \sqrt{2gh_1} + \frac{5}{2}g \quad \text{(5)} \]

### Step 4: Equate equations (3) and (5)
From equations (3) and (5), we can set them equal:
\[ 2\sqrt{2gh_1} + 2g = \sqrt{2gh_1} + \frac{5}{2}g \]
Rearranging gives:
\[ 2\sqrt{2gh_1} - \sqrt{2gh_1} = \frac{5}{2}g - 2g \]
This simplifies to:
\[ \sqrt{2gh_1} = \frac{1}{2}g \]
Squaring both sides:
\[ 2gh_1 = \frac{1}{4}g^2 \quad \Rightarrow \quad h_1 = \frac{g}{8} \quad \text{(6)} \]

### Step 5: Find the time taken from A to B
Substituting \( h_1 \) back into equation (1):
\[ h_1 = \frac{1}{2}g t^2 \quad \Rightarrow \quad \frac{g}{8} = \frac{1}{2}g t^2 \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{8} = \frac{1}{2}t^2 \]
Multiplying both sides by 2:
\[ \frac{1}{4} = t^2 \]
Taking the square root:
\[ t = \frac{1}{2} \text{ seconds} \]

### Final Answer
The time taken to move from A to B is \( 0.5 \) seconds.

---