A juggler maintain 10 balls in motion, making each of them to rise at a height of 80 m from his hands. To keep proper distance between them, the time interval maintained by juggler would be

Options

A. 0.6 s
B. 0.8 s
C. 1.0 s
D. 1.2 s

Answer

To solve the problem of how much time interval a juggler should maintain between throwing 10 balls to reach a height of 80 meters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height (h) = 80 m
- Number of balls (n) = 10
- Acceleration due to gravity (g) = 10 m/s² (approximated)

2. **Calculate the Initial Velocity (u):**
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Here, at the maximum height, the final velocity (v) is 0. Thus, we can rearrange the equation:
\[ 0 = u^2 - 2gs \]
Solving for u gives:
\[ u^2 = 2gs \implies u = \sqrt{2gs} \]
Substituting the values:
\[ u = \sqrt{2 \times 10 \times 80} = \sqrt{1600} = 40 \text{ m/s} \]

3. **Calculate the Time to Reach Maximum Height:**
We can use another kinematic equation:
\[ v = u + at \]
Rearranging for time (t):
\[ t = \frac{v - u}{a} = \frac{0 - 40}{-10} = 4 \text{ seconds} \]
This is the time taken to reach the maximum height.

4. **Calculate the Total Time for One Ball:**
The total time for one ball to go up and come back down is:
\[ \text{Total time} = \text{Time up} + \text{Time down} = 4 \text{ seconds} + 4 \text{ seconds} = 8 \text{ seconds} \]

5. **Calculate the Time Interval Between Throws:**
Since the juggler is juggling 10 balls, the time interval (T) between throwing each ball is:
\[ T = \frac{\text{Total time for one ball}}{\text{Number of balls}} = \frac{8 \text{ seconds}}{10} = 0.8 \text{ seconds} \]

### Final Answer:
The time interval maintained by the juggler would be **0.8 seconds**.

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