Two particles A and B are dropped from the heights of 5 m and 20 m respectively. Then the ratio of time taken by A to that taken by B, to reach the ground is

Options

A. 1:4
B. 2:1
C. 1:2
D. 1:1

Answer

To solve the problem of finding the ratio of the time taken by particles A and B to reach the ground when dropped from heights of 5 m and 20 m respectively, we can follow these steps:

### Step 1: Understand the problem
We have two particles:
- Particle A is dropped from a height \( h_1 = 5 \, \text{m} \)
- Particle B is dropped from a height \( h_2 = 20 \, \text{m} \)

Both particles are dropped from rest, meaning their initial velocity \( u = 0 \).

### Step 2: Use the equation of motion
The equation of motion for an object in free fall is given by:
\[ h = ut + \frac{1}{2}gt^2 \]
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ h = \frac{1}{2}gt^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 3: Calculate time for particle A
For particle A:
\[ h_1 = \frac{1}{2}gt_1^2 \]
Substituting \( h_1 = 5 \, \text{m} \):
\[ 5 = \frac{1}{2}gt_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{10}{g} \]
Taking the square root:
\[ t_1 = \sqrt{\frac{10}{g}} \]

### Step 4: Calculate time for particle B
For particle B:
\[ h_2 = \frac{1}{2}gt_2^2 \]
Substituting \( h_2 = 20 \, \text{m} \):
\[ 20 = \frac{1}{2}gt_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{40}{g} \]
Taking the square root:
\[ t_2 = \sqrt{\frac{40}{g}} \]

### Step 5: Find the ratio of times
Now, we need to find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{10}{g}}}{\sqrt{\frac{40}{g}}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{\sqrt{10}}{\sqrt{40}} = \frac{\sqrt{10}}{\sqrt{4 \cdot 10}} = \frac{\sqrt{10}}{2\sqrt{10}} = \frac{1}{2} \]

### Final Answer
Thus, the ratio of the time taken by A to that taken by B to reach the ground is:
\[ \frac{t_1}{t_2} = \frac{1}{2} \]