A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is

Options

A. `g/(2(n-1)^2)`
B. `g/(2n^2)`
C. `g/n^2`
D. `g/n`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
A boy throws \( n \) balls per second at regular intervals. When the first ball reaches its maximum height, he throws the second ball. We need to find the maximum height reached by each ball.

### Step 2: Determine the Time Interval Between Throws
Since the boy throws \( n \) balls in one second, the time interval \( t \) between two successive throws is given by:
\[ t = \frac{1}{n} \text{ seconds} \]

### Step 3: Use Kinematic Equations
To find the maximum height reached by the balls, we need to determine the initial velocity \( u \) of the balls. At maximum height, the final velocity \( v \) of the ball is \( 0 \) m/s. The acceleration \( a \) acting on the ball is due to gravity, which is \( -g \) (acting downwards).

Using the kinematic equation:
\[ v = u + at \]
Substituting the known values:
\[ 0 = u - gt \]
Rearranging gives:
\[ u = gt \]

### Step 4: Substitute the Time Interval
We know from Step 2 that \( t = \frac{1}{n} \). Substituting this into the equation for \( u \):
\[ u = g \left(\frac{1}{n}\right) = \frac{g}{n} \]

### Step 5: Calculate the Maximum Height
The formula for the maximum height \( h_{\text{max}} \) reached by a ball thrown vertically is:
\[ h_{\text{max}} = \frac{u^2}{2g} \]
Substituting \( u = \frac{g}{n} \):
\[ h_{\text{max}} = \frac{\left(\frac{g}{n}\right)^2}{2g} \]
This simplifies to:
\[ h_{\text{max}} = \frac{g^2}{2n^2g} = \frac{g}{2n^2} \]

### Final Answer
Thus, the maximum height reached by each ball is:
\[ h_{\text{max}} = \frac{g}{2n^2} \]