A ball of mass `m_(1)` and another ball of mass `m_(2)` are dropped from equal height. If the time taken by the balls are `t_(1)` and `t_(2)`, respectively, then

Options

A. `t_(1)=t_(2)`
B. `t_(1)=2t_(2)`
C. `t_(1)/t_(2)=m_(1)/m_(2)`
D. `t_(1)/t_(2)=m_(2)/m_(1)`

Explanation

To solve the problem, we need to analyze the motion of two balls dropped from the same height. We will use the equations of motion to derive the relationship between the times taken by the two balls.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Two balls with masses \( m_1 \) and \( m_2 \) are dropped from the same height \( h \).
- The time taken by ball 1 is \( t_1 \) and by ball 2 is \( t_2 \).
- Both balls are dropped from rest, so their initial velocities \( u = 0 \).

2. **Use the Equation of Motion:**
- The equation of motion for an object in free fall is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the displacement (height \( h \)), \( u \) is the initial velocity (which is 0), \( a \) is the acceleration (which is \( g \), the acceleration due to gravity), and \( t \) is the time taken.

3. **Apply the Equation for Both Balls:**
- For ball 1:
\[ h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \implies h = \frac{1}{2} g t_1^2 \]
- For ball 2:
\[ h = 0 \cdot t_2 + \frac{1}{2} g t_2^2 \implies h = \frac{1}{2} g t_2^2 \]

4. **Set the Equations Equal to Each Other:**
- Since both balls are dropped from the same height \( h \), we can set the two equations equal:
\[ \frac{1}{2} g t_1^2 = \frac{1}{2} g t_2^2 \]

5. **Simplify the Equation:**
- We can cancel \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ t_1^2 = t_2^2 \]

6. **Take the Square Root:**
- Taking the square root of both sides gives:
\[ t_1 = t_2 \]

### Conclusion:
The time taken by both balls to reach the ground is the same, regardless of their masses. Therefore, the relationship is:
\[ t_1 = t_2 \]