A particle is released from rest from a tower of height 3h. The ratio of time intervals for fall of equal height h i.e. `t_(1):t_(2):t_(3)` is :

Options

A. `5:3:1`
B. `3:2:1`
C. `9:4:1`
D. `1:(sqrt(2)-1):(sqrt(3):sqrt(2))`

Answer

To solve the problem of finding the ratio of time intervals \( t_1 : t_2 : t_3 \) for a particle falling from a height of \( 3h \) in equal segments of height \( h \), we can follow these steps:

### Step 1: Define the heights and time intervals
The particle is released from a height of \( 3h \). We can divide this height into three equal parts:
- From point A (height = 3h) to point B (height = 2h): height = \( h \)
- From point B (height = 2h) to point C (height = h): height = \( h \)
- From point C (height = h) to point D (height = 0): height = \( h \)

Let:
- \( t_1 \) = time taken to fall from A to B
- \( t_2 \) = time taken to fall from B to C
- \( t_3 \) = time taken to fall from C to D

### Step 2: Calculate \( t_1 \)
Using the second equation of motion for the first segment (A to B):
\[ s = ut + \frac{1}{2}gt^2 \]
Here, \( s = h \), \( u = 0 \) (initial velocity), and \( g \) is the acceleration due to gravity. Thus:
\[ h = 0 + \frac{1}{2}g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2h}{g} \]
Taking the square root:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

### Step 3: Calculate \( t_2 \)
For the second segment (B to C), the particle has already fallen \( h \) from A to B, so it starts from rest at height \( 2h \). The total distance fallen from A to C is \( 2h \):
\[ 2h = 0 + \frac{1}{2}g t_{AC}^2 \]
Rearranging gives:
\[ t_{AC}^2 = \frac{4h}{g} \]
Taking the square root:
\[ t_{AC} = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}} \]
Now, to find \( t_2 \):
\[ t_2 = t_{AC} - t_1 = 2\sqrt{\frac{h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{h}{g}}(2 - \sqrt{2}) \]

### Step 4: Calculate \( t_3 \)
For the third segment (C to D), the total distance fallen from A to D is \( 3h \):
\[ 3h = 0 + \frac{1}{2}g t_{AD}^2 \]
Rearranging gives:
\[ t_{AD}^2 = \frac{6h}{g} \]
Taking the square root:
\[ t_{AD} = \sqrt{\frac{6h}{g}} \]
Now, to find \( t_3 \):
\[ t_3 = t_{AD} - t_{AC} = \sqrt{\frac{6h}{g}} - 2\sqrt{\frac{h}{g}} = \sqrt{\frac{h}{g}}(\sqrt{6} - 2) \]

### Step 5: Find the ratio \( t_1 : t_2 : t_3 \)
Now we have:
- \( t_1 = \sqrt{\frac{2h}{g}} \)
- \( t_2 = \sqrt{\frac{h}{g}}(2 - \sqrt{2}) \)
- \( t_3 = \sqrt{\frac{h}{g}}(\sqrt{6} - 2) \)

To find the ratio, we can express all terms in terms of \( \sqrt{\frac{h}{g}} \):
\[ t_1 : t_2 : t_3 = \sqrt{2} : (2 - \sqrt{2}) : (\sqrt{6} - 2) \]

### Final Ratio
Thus, the final ratio of time intervals is:
\[ t_1 : t_2 : t_3 = \sqrt{2} : (2 - \sqrt{2}) : (\sqrt{6} - 2) \]