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

Answer

To solve the problem, we will go through the following steps:

### Step 1: Understand the Time Interval Between Throws
The boy throws \( n \) balls per second at regular time intervals. Therefore, the time interval \( t \) between two successive throws is given by:
\[ t = \frac{1}{n} \]

**Hint:** Remember that if \( n \) balls are thrown in one second, the time between each throw is the reciprocal of \( n \).

### Step 2: Determine the Time to Reach Maximum Height
When the first ball is thrown, it takes a certain amount of time to reach its maximum height. This time is equal to the time interval \( t \) since the second ball is thrown when the first ball reaches its maximum height. Thus, the time taken \( t \) to reach maximum height is:
\[ t = \frac{1}{n} \]

**Hint:** The time to reach maximum height is the same as the time interval between throws.

### Step 3: Use Kinematic Equations to Find Initial Velocity
At the maximum height, the final velocity \( v \) of the ball is \( 0 \). We can use the kinematic equation:
\[ v = u - gt \]
Where:
- \( v \) = final velocity (0 at maximum height)
- \( u \) = initial velocity (velocity of projection)
- \( g \) = acceleration due to gravity
- \( t \) = time taken to reach maximum height

Substituting \( v = 0 \) and \( t = \frac{1}{n} \):
\[ 0 = u - g \left(\frac{1}{n}\right) \]
Rearranging gives:
\[ u = g \left(\frac{1}{n}\right) = \frac{g}{n} \]

**Hint:** Use the fact that the final velocity at the maximum height is zero to find the initial velocity.

### Step 4: Calculate the Maximum Height
The maximum height \( h \) reached by the ball can be calculated using the formula:
\[ h = \frac{u^2}{2g} \]
Substituting the value of \( u \):
\[ h = \frac{\left(\frac{g}{n}\right)^2}{2g} \]
This simplifies to:
\[ h = \frac{g^2}{n^2 \cdot 2g} = \frac{g}{2n^2} \]

**Hint:** Remember to square the initial velocity and divide by \( 2g \) to find the maximum height.

### Step 5: Conclusion
Thus, the maximum height reached by each ball is:
\[ h = \frac{g}{2n^2} \]
This matches with option (b) from the given choices.

**Final Answer:** The maximum height reached by each ball is \( \frac{g}{2n^2} \).