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 use the equations of motion under uniform acceleration due to gravity.

### Step 1: Understand the problem
The boy throws a ball into the air every 2 seconds, and the next ball is thrown when the first ball reaches its maximum height (where its velocity becomes zero).

### Step 2: Determine the time of ascent
Since the boy throws the ball every 2 seconds and the second ball is thrown when the first ball reaches its maximum height, the time taken for the first ball to reach its maximum height is 2 seconds.

### Step 3: Use the equation of motion
At maximum height, the final velocity (v) of the ball is 0 m/s. We can use the following equation of motion to find the initial velocity (u) of the ball:

\[ v = u - gt \]

Where:
- \( v = 0 \) (final velocity at maximum height)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 2 \, \text{s} \) (time of ascent)

Substituting the values into the equation:

\[ 0 = u - (9.8)(2) \]

### Step 4: Solve for initial velocity (u)
Rearranging the equation gives:

\[ u = 9.8 \times 2 \]
\[ u = 19.6 \, \text{m/s} \]

### Step 5: Calculate the maximum height (h)
Now, we can use the formula for maximum height reached by the ball:

\[ h = \frac{u^2}{2g} \]

Substituting the values we found:

\[ h = \frac{(19.6)^2}{2 \times 9.8} \]

### Step 6: Calculate the value
Calculating \( (19.6)^2 \):

\[ (19.6)^2 = 384.16 \]

Now substituting this value into the height formula:

\[ h = \frac{384.16}{19.6} \]
\[ h = 19.6 \, \text{m} \]

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

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