Question

a juggler keeps on moving four balls in the 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 position of other balls from his hand will be (in m)
40,80,40
60,80,60
40,60,80
80,60,80​

Answer 1

Answer:

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