A boy throws balls into air at regular interval of 2 second. The next ball is thrown when the velocity of first ball is zero. How high do the ball rise above his hand? [Take `g=9.8m//s^(2)`]

Options

A. 4.9 m
B. 9.8 m
C. 19.6 m
D. 29.4 m

Answer

To solve the problem step by step, we will follow the concepts of kinematics in motion under gravity.

### Step 1: Understand the Problem
The boy throws balls into the air at regular intervals of 2 seconds. The next ball is thrown when the first ball reaches its maximum height, which is when its velocity becomes zero.

### Step 2: Determine the Initial Velocity
When the first ball is thrown, it will take 2 seconds to reach its maximum height (where its velocity is zero). We can use the formula for the time taken to reach maximum height:
\[ t = \frac{u}{g} \]
Where:
- \( t \) is the time taken to reach maximum height (2 seconds)
- \( u \) is the initial velocity of the ball
- \( g \) is the acceleration due to gravity (9.8 m/s²)

Substituting the known values:
\[ 2 = \frac{u}{9.8} \]

### Step 3: Solve for Initial Velocity \( u \)
Rearranging the equation to solve for \( u \):
\[ u = 2g = 2 \times 9.8 = 19.6 \, \text{m/s} \]

### Step 4: Calculate the Maximum Height
Now that we have the initial velocity, we can calculate the maximum height \( h \) the ball reaches using the formula:
\[ h = \frac{u^2}{2g} \]
Substituting the value of \( u \):
\[ h = \frac{(19.6)^2}{2 \times 9.8} \]

### Step 5: Simplify the Calculation
Calculating \( (19.6)^2 \):
\[ (19.6)^2 = 384.16 \]
Now substituting this back into the height formula:
\[ h = \frac{384.16}{19.6} \]

### Step 6: Final Calculation
Now, simplifying:
\[ h = 19.6 \, \text{m} \]

Thus, the ball rises to a height of **19.6 meters** above the boy's hand.

### Summary of the Solution
The maximum height the ball rises above the boy's hand is **19.6 meters**.

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