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Question

A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand (speed =20 m s1), the position of other balls (height in m) will be (table g= 10 m s2)


A

10, 20, 10

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B

15, 20, 15

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C

5, 15, 20

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D

5, 10, 20

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Solution

The correct option is D. 15, 20, 15.

The time taken by the ball to reach its maximum height is given by the first equation of motion as,

0 = 20 - 10t = 2 sec.

So, the total time will be 4 sec i.e. juggler will touch the same ball after 4 sec only. Since he is juggling 4 balls at regular intervals, i.e. each of them is spaced at 1 sec distance interval.

Therefore the position of balls are given by the second equation of motion as follows,

h3 = 20 × 1 - 12 × 10 (1)2 = 15 m

h2 = 20 × 2 - 12 × 10 (2)2 = 20 m

h1 = 20 × 3 - 12 × 10 (3)2 = 15 m


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