A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand `("speed" = 20 ms^-1)` the positions of other balls (height in m) `("Take" g = 10 ms^-2)`.

Options

A. `10,20,10`
B. `15, 20,15`
C. `5, 15,20`
D. `5,10,20`

Explanation

To solve the problem of the juggler throwing four balls in the air, we will follow these steps:

### Step 1: Understand the motion of the balls
When the juggler throws the balls, they are thrown at a speed of 20 m/s. The balls will rise until they reach their maximum height, where their velocity will become 0, and then they will fall back down due to gravity.

### Step 2: Calculate the time taken for one ball to reach its maximum height
Using the first equation of motion:
\[ v = u + at \]
Where:
- \( v = 0 \) m/s (final velocity at the maximum height)
- \( u = 20 \) m/s (initial velocity)
- \( a = -g = -10 \) m/s² (acceleration due to gravity)

Rearranging the equation gives:
\[ 0 = 20 - 10t \]
\[ 10t = 20 \]
\[ t = 2 \text{ seconds} \]

### Step 3: Calculate the total time for the ball to return to the juggler's hand
The total time for the ball to go up and come back down is:
\[ T = t_{up} + t_{down} = 2 + 2 = 4 \text{ seconds} \]

### Step 4: Determine the time interval between throws
Since the juggler throws four balls in total and the total time for the first ball to return is 4 seconds, the time interval between each throw is:
\[ \text{Time interval} = \frac{T}{\text{Number of balls}} = \frac{4}{4} = 1 \text{ second} \]

### Step 5: Calculate the heights of the balls at the moment the fourth ball is thrown
1. **Height of the 1st ball (thrown 3 seconds before the 4th ball)**:
\[ h_1 = ut + \frac{1}{2}at^2 \]
\[ h_1 = 20 \times 3 + \frac{1}{2}(-10)(3^2) \]
\[ h_1 = 60 - 45 = 15 \text{ m} \]

2. **Height of the 2nd ball (thrown 2 seconds before the 4th ball)**:
\[ h_2 = ut + \frac{1}{2}at^2 \]
\[ h_2 = 20 \times 2 + \frac{1}{2}(-10)(2^2) \]
\[ h_2 = 40 - 20 = 20 \text{ m} \]

3. **Height of the 3rd ball (thrown 1 second before the 4th ball)**:
\[ h_3 = ut + \frac{1}{2}at^2 \]
\[ h_3 = 20 \times 1 + \frac{1}{2}(-10)(1^2) \]
\[ h_3 = 20 - 5 = 15 \text{ m} \]

### Step 6: Summary of heights
- Height of the 1st ball: 15 m
- Height of the 2nd ball: 20 m
- Height of the 3rd ball: 15 m

### Final Answer
The positions of the balls when the fourth ball is thrown are:
- 1st ball: 15 m
- 2nd ball: 20 m
- 3rd ball: 15 m