Step 1: Understanding the Concept:
The problem states that a new ball is thrown when the previous one reaches its maximum height. This means the time taken for a ball to reach its maximum height is equal to the time interval between two consecutive throws. We can then use kinematic equations to relate this time to the maximum height.
Step 2: Key Formula or Approach:
1. Find the time interval between throws.
2. This time interval is the time of ascent ($t_{up}$) for one ball.
3. At the maximum height, the final velocity ($v$) of the ball is 0.
4. Use the kinematic equations $v = u + at$ and $v^2 = u^2 + 2as$ to find the maximum height ($s=H$). Let's use upward as positive, so $a = -g$.
Step 3: Detailed Explanation:
The juggler throws 'n' balls each second. This means the time interval ($\Delta t$) between two consecutive throws is:
\[ \Delta t = \frac{1}{n} \text{ seconds} \]
According to the problem, this is the time it takes for a ball to reach its maximum height.
So, the time of ascent, $t_{up} = \frac{1}{n}$.
At the maximum height, the vertical velocity is zero ($v=0$). We can find the initial velocity ($u$) using $v = u + at$:
\[ 0 = u + (-g)t_{up} \]
\[ u = g t_{up} = g \left(\frac{1}{n}\right) = \frac{g}{n} \]
Now we can find the maximum height ($H$) using the equation $v^2 = u^2 + 2as$:
\[ 0^2 = u^2 + 2(-g)H \]
\[ u^2 = 2gH \]
\[ H = \frac{u^2}{2g} \]
Substitute the expression for the initial velocity $u$:
\[ H = \frac{(\frac{g}{n})^2}{2g} = \frac{\frac{g^2}{n^2}}{2g} = \frac{g^2}{2gn^2} = \frac{g}{2n^2} \]
Step 4: Final Answer:
The maximum height the balls rise to is $\frac{g}{2n^2}$. Therefore, option (A) is correct.