A man throw rings into air one after the other at an interval of one second. The next ring is thrown when the velocity of the ring thrown earlier is zero. The height to which the ring rise is `("take" g = 10ms^(-2))`

Options

A. 15 m
B. 5 m
C. 10 m
D. 20 m

Explanation

To solve the problem step-by-step, let's analyze the situation described:

1. **Understanding the Problem**:
- A man throws rings into the air at intervals of one second.
- Each ring is thrown when the previous ring reaches its maximum height, where its velocity becomes zero.

2. **Identifying Key Variables**:
- Let the initial velocity of the ring (when thrown) be \( u \).
- The time taken to reach maximum height is \( t \).
- The acceleration due to gravity \( g \) is given as \( 10 \, \text{m/s}^2 \).

3. **Using the Kinematic Equation**:
- The kinematic equation that relates height, initial velocity, time, and acceleration is:
\[ h = ut - \frac{1}{2}gt^2 \]
- At the maximum height, the final velocity \( v = 0 \).

4. **Finding Time to Reach Maximum Height**:
- The time taken to reach maximum height can be determined using the equation:
\[ v = u - gt \]
- Setting \( v = 0 \) (at maximum height):
\[ 0 = u - gt \implies u = gt \]
- Since the next ring is thrown after 1 second, we can assume \( t = 1 \, \text{s} \):
\[ u = g \cdot 1 = 10 \, \text{m/s} \]

5. **Calculating the Height**:
- Now, substituting \( u = 10 \, \text{m/s} \) and \( t = 1 \, \text{s} \) into the height equation:
\[ h = ut - \frac{1}{2}gt^2 \]
\[ h = 10 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 \]
\[ h = 10 - 5 = 5 \, \text{m} \]

6. **Conclusion**:
- The height to which the ring rises is \( 5 \, \text{m} \).

### Final Answer:
The height to which the ring rises is **5 meters**.