Question

A juggler keeps on moving four balls in air by throwing the balls after regular intervals. He throws the ball with a speed of 40 m/s. When one ball leaves
his hand, the positions of other balls from his hand will be (in m)​

Answer 1

Explanation:

Time taken by the small ball to return to the hands of the juggler is

g

2v

=

10

2×20

=4s. So, he is throwing the balls after 1s each. Let at some instant, he throws the ball number 4. Before 1s of throwing it, he throws ball 3. So, the height of ball 3 is

h=ut−

2

1

gt

2

h

3

=20×1−

2

1

(10)(1)

2

=15m

Before 2s, he throws ball 2. So, the height of ball 2 is

h

2

=20×2−

2

1

×10(2)

2

=20m

Before 3s, he throws ball 1. So, the height of ball 1 is

h

1

=20×3−

2

1

×10(3)

2

⇒h

1

=15m

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