Body A of mass M is dropped from a height of 1 m and body. B of mass 3M is dropped from a height of 9 m. The ratio of time taken by the bodies 1 and 2 to reach the ground is

Options

A. `1:1`
B. `1:3`
C. `3:1`
D. `9:1`

Answer

To solve the problem, we need to find the time taken by two bodies, A and B, to reach the ground when dropped from different heights. Let's break down the solution step by step.

### Step 1: Identify the given data
- Body A (mass = M) is dropped from a height \( h_A = 1 \, \text{m} \).
- Body B (mass = 3M) is dropped from a height \( h_B = 9 \, \text{m} \).
- Both bodies are dropped from rest, so their initial velocities \( u_A = 0 \) and \( u_B = 0 \).
- The acceleration due to gravity \( g \) is constant for both bodies.

### Step 2: Use the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance fallen,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time taken.

### Step 3: Calculate time for Body A
For Body A:
- \( s = h_A = 1 \, \text{m} \)
- \( u_A = 0 \)
- \( a = g \)

Substituting into the equation:
\[ 1 = 0 \cdot t_A + \frac{1}{2} g t_A^2 \]
This simplifies to:
\[ 1 = \frac{1}{2} g t_A^2 \]
Rearranging gives:
\[ t_A^2 = \frac{2}{g} \]
Taking the square root:
\[ t_A = \sqrt{\frac{2}{g}} \]

### Step 4: Calculate time for Body B
For Body B:
- \( s = h_B = 9 \, \text{m} \)
- \( u_B = 0 \)
- \( a = g \)

Substituting into the equation:
\[ 9 = 0 \cdot t_B + \frac{1}{2} g t_B^2 \]
This simplifies to:
\[ 9 = \frac{1}{2} g t_B^2 \]
Rearranging gives:
\[ t_B^2 = \frac{18}{g} \]
Taking the square root:
\[ t_B = \sqrt{\frac{18}{g}} = \sqrt{\frac{9 \cdot 2}{g}} = 3\sqrt{\frac{2}{g}} \]

### Step 5: Find the ratio of time taken
Now, we need to find the ratio of the times taken by A and B:
\[ \frac{t_A}{t_B} = \frac{\sqrt{\frac{2}{g}}}{3\sqrt{\frac{2}{g}}} \]
The \( \sqrt{\frac{2}{g}} \) terms cancel out:
\[ \frac{t_A}{t_B} = \frac{1}{3} \]

### Step 6: Conclusion
Thus, the ratio of the time taken by body A to the time taken by body B to reach the ground is:
\[ \text{Ratio } (t_A : t_B) = 1 : 3 \]