A particle is released from rest from a tower of height 3h. The ratio of times to fall equal heights 'h' i.e. `t_1:t_2:t_3` is (`sqrt3` =1.17 & `sqrt2` = 1.4)

Options

A. `sqrt3:sqrt2:1`
B. `3:2:1`
C. `1:14:1.7`
D. `10:4:3`

Answer

To solve the problem step by step, we need to find the ratio of times \(T_1 : T_2 : T_3\) taken by a particle falling from a height of \(3h\) to fall equal heights of \(h\).

### Step 1: Understanding the Problem
The particle is released from rest from a height of \(3h\). We need to find the times taken to fall the first height \(h\) (denote this time as \(T_1\)), the next height \(h\) (denote this time as \(T_2\)), and the last height \(h\) (denote this time as \(T_3\)).

### Step 2: Using the Kinematic Equation
The kinematic equation for displacement when an object is in free fall is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the particle is released from rest, the initial velocity \(u = 0\), and the acceleration \(a = g\) (acceleration due to gravity). Thus, the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]

### Step 3: Finding \(T_1\)
For the first height \(h\):
\[ h = \frac{1}{2} g T_1^2 \]
Rearranging this gives:
\[ T_1^2 = \frac{2h}{g} \]
Taking the square root:
\[ T_1 = \sqrt{\frac{2h}{g}} \]

### Step 4: Finding \(T_2\)
For the second height \(h\) (from \(h\) to \(2h\)), the total distance fallen is \(2h\):
\[ 2h = \frac{1}{2} g (T_1 + T_2)^2 \]
Substituting \(T_1\):
\[ 2h = \frac{1}{2} g \left(\sqrt{\frac{2h}{g}} + T_2\right)^2 \]
Expanding and simplifying:
\[ 2h = \frac{1}{2} g \left(\frac{2h}{g} + 2\sqrt{\frac{2h}{g}} T_2 + T_2^2\right) \]
This gives:
\[ 4h = 2h + g \sqrt{\frac{2h}{g}} T_2 + \frac{1}{2} g T_2^2 \]
Solving for \(T_2\):
\[ T_2 = \sqrt{\frac{2(2h)}{g}} - T_1 \]
\[ T_2 = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{2} - 1) \]

### Step 5: Finding \(T_3\)
For the third height \(h\) (from \(2h\) to \(3h\)), the total distance fallen is \(3h\):
\[ 3h = \frac{1}{2} g (T_1 + T_2 + T_3)^2 \]
Substituting the values of \(T_1\) and \(T_2\):
\[ 3h = \frac{1}{2} g \left(T_1 + T_2 + T_3\right)^2 \]
Expanding and simplifying gives:
\[ 6h = g \left(T_1 + T_2 + T_3\right)^2 \]
Solving for \(T_3\):
\[ T_3 = \sqrt{\frac{6h}{g}} - (T_1 + T_2) \]

### Step 6: Finding the Ratios
Now we have:
- \(T_1 = \sqrt{\frac{2h}{g}}\)
- \(T_2 = \sqrt{\frac{2h}{g}}(\sqrt{2} - 1)\)
- \(T_3 = \sqrt{\frac{2h}{g}}(\sqrt{3} - \sqrt{2})\)

The ratio \(T_1 : T_2 : T_3\) can be expressed as:
\[ 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2}) \]

### Final Step: Substituting Values
Using the approximations \(\sqrt{2} \approx 1.4\) and \(\sqrt{3} \approx 1.17\):
- \(T_1 : T_2 : T_3 \approx 1 : (1.4 - 1) : (1.17 - 1.4)\)
- This gives \(1 : 0.4 : 0.3\)

### Conclusion
Thus, the final ratio is:
\[ T_1 : T_2 : T_3 = 10 : 4 : 3 \]