Question:

A juggler keeps on moving four balls in air throwing the balls after regular intervals. When one ball leaves his hand (speed = \(20\text{ ms}^{-1}\)), the position of other balls (height in metre) will be (take \(g = 10\text{ ms}^{-2}\))

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The juggler's hand is at height 0 when ball leaves; heights are measured from hand level.
Updated On: Apr 20, 2026
  • 10, 20, 10
  • 15, 20, 15
  • 5, 15, 20
  • 5, 10, 20
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Time of flight for one ball: \(T = \frac{2u}{g} = \frac{2\times20}{10} = 4 \text{ s}\). Balls are thrown at regular intervals of 1 second.
Step 2: Detailed Explanation:
At \(t=0\), one ball is thrown; previous balls have been in air for 1 s, 2 s, and 3 s. Height formula: \(h = ut - \frac{1}{2}gt^2\). For \(t=1\): \(h = 20\times1 - 5\times1 = 15 \text{ m}\). For \(t=2\): \(h = 40 - 20 = 20 \text{ m}\). For \(t=3\): \(h = 60 - 45 = 15 \text{ m}\). Heights are: 15 m, 20 m, 15 m.
Step 3: Final Answer:
\[ \boxed{15,\; 20,\; 15} \]
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