A juggler maintains four balls in motion, making each to them to rise a height of `20 m` from his hand. What time interval should he maintain, for the alphaer distance between them.

Options

A. 3 s
B. `(3)/(2) s`
C. 1 s
D. 2 s

Explanation

To solve the problem of finding the time interval a juggler should maintain between throwing four balls that rise to a height of 20 meters, we can follow these steps:

### Step 1: Determine the time taken to reach the maximum height

When a ball is thrown upwards, it will rise to a maximum height before falling back down. The time taken to reach the maximum height can be calculated using the kinematic equation:

\[ h = \frac{1}{2} g t^2 \]

where:
- \( h \) is the height (20 m),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time taken to reach the maximum height.

Rearranging the equation to solve for \( t \):

\[ t = \sqrt{\frac{2h}{g}} \]

### Step 2: Substitute the values

Substituting the values into the equation:

\[ t = \sqrt{\frac{2 \times 20 \, \text{m}}{9.81 \, \text{m/s}^2}} \]

Calculating the value:

\[ t = \sqrt{\frac{40}{9.81}} \approx \sqrt{4.08} \approx 2.02 \, \text{s} \]

### Step 3: Determine the total time for the ball to go up and come down

Since the ball takes the same amount of time to come down as it does to go up, the total time for one complete cycle (up and down) is:

\[ T = 2t \approx 2 \times 2.02 \, \text{s} \approx 4.04 \, \text{s} \]

### Step 4: Calculate the time interval between throws

To maintain four balls in motion, the juggler needs to throw each ball at intervals that allow them to be at different heights at the same time. Since there are four balls, the time interval between each throw should be:

\[ \Delta t = \frac{T}{4} = \frac{4.04 \, \text{s}}{4} \approx 1.01 \, \text{s} \]

### Conclusion

The juggler should maintain an interval of approximately **1.01 seconds** between each throw of the balls to keep them at proper distances.

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