A juggler throws ball into air. He throus one whenever the previus one is at its highest point. How high do the balls rise if he throus (n) balls each second. Acceleration the to gravity=g`.

Options

A. `xv_(x)+yv_(y) lt 0`
B. `xv_(x)+yv_(y) gt 0`
C. `a_(x)v_(x)+a_(y)v_(y) lt 0`
D. `a_(x)v_(x)+a_(y)v_(y) gt 0`

Explanation

To solve the problem of how high the balls rise when a juggler throws \( n \) balls each second, we can break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Time Interval**:
- The juggler throws \( n \) balls each second. This means that he throws one ball every \( \frac{1}{n} \) seconds.

2. **Time to Reach Maximum Height**:
- Each ball takes \( \frac{1}{n} \) seconds to reach its maximum height. This is the same time interval in which the juggler throws the next ball.

3. **Using the First Equation of Motion**:
- At the maximum height, the final velocity \( v = 0 \). We can use the first equation of motion:
\[ v = u - g t \]
- Here, \( u \) is the initial velocity of the ball, \( g \) is the acceleration due to gravity, and \( t \) is the time taken to reach the maximum height, which is \( \frac{1}{n} \) seconds.

4. **Finding the Initial Velocity**:
- Rearranging the equation for \( u \):
\[ 0 = u - g \left(\frac{1}{n}\right) \]
\[ u = g \left(\frac{1}{n}\right) \]

5. **Calculating the Maximum Height**:
- We can now use the second equation of motion to find the maximum height \( h \):
\[ h = ut - \frac{1}{2} g t^2 \]
- Substituting \( u \) and \( t \):
\[ h = \left(g \frac{1}{n}\right) \left(\frac{1}{n}\right) - \frac{1}{2} g \left(\frac{1}{n}\right)^2 \]
- Simplifying this expression:
\[ h = \frac{g}{n^2} - \frac{1}{2} g \frac{1}{n^2} \]
\[ h = \frac{g}{2n^2} \]

6. **Final Result**:
- Therefore, the maximum height reached by each ball is:
\[ h = \frac{g}{2n^2} \]

### Summary:
The maximum height reached by the balls when the juggler throws \( n \) balls each second is \( \frac{g}{2n^2} \).