Two balls A & B, mass of A is m and that of B is `5` m are dropped from the towers of height `36` m and `64` m respectively. The ratio of the time taken by them to reach the ground is :

Options

A. `0.75`
B. `3//4`
C. Both (a) and (b)
D. `5//2`

Explanation

To solve the problem of finding the ratio of the time taken by two balls A and B to reach the ground when dropped from different heights, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the heights from which the balls are dropped:**
- Ball A is dropped from a height \( H_A = 36 \, \text{m} \).
- Ball B is dropped from a height \( H_B = 64 \, \text{m} \).

2. **Use the kinematic equation for free fall:**
The equation for the distance covered under uniform acceleration (due to gravity) is given by:
\[ S = Ut + \frac{1}{2} a t^2 \]
where:
- \( S \) is the distance (height of the tower),
- \( U \) is the initial velocity (which is 0 for both balls since they are dropped),
- \( a \) is the acceleration due to gravity (denoted as \( g \)),
- \( t \) is the time taken to reach the ground.

Since the initial velocity \( U = 0 \), the equation simplifies to:
\[ S = \frac{1}{2} g t^2 \]

3. **Rearranging the equation:**
We can rearrange the equation to express time \( t \):
\[ t^2 = \frac{2S}{g} \]
Therefore, the time taken for each ball can be expressed as:
\[ t_A = \sqrt{\frac{2H_A}{g}} \quad \text{and} \quad t_B = \sqrt{\frac{2H_B}{g}} \]

4. **Find the ratio of the times:**
We need to find the ratio \( \frac{t_A}{t_B} \):
\[ \frac{t_A}{t_B} = \frac{\sqrt{\frac{2H_A}{g}}}{\sqrt{\frac{2H_B}{g}}} \]
This simplifies to:
\[ \frac{t_A}{t_B} = \sqrt{\frac{H_A}{H_B}} \]

5. **Substituting the heights:**
Now, substituting the values of \( H_A \) and \( H_B \):
\[ \frac{t_A}{t_B} = \sqrt{\frac{36}{64}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

6. **Final Result:**
Thus, the ratio of the time taken by balls A and B to reach the ground is:
\[ \frac{t_A}{t_B} = \frac{3}{4} \]