Manipal Entrance Test 2015 Physics Paper

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

2v

g

=

2×20

10

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

1

2

gt2
h3=20×1−

1

2

(10)(1)2=15m
Before 2 s, he throws ball 2. So, the height of ball 2 is
h2=20×2−

1

2

×10(2)2=20m
Before 3 s, he throws ball 1. So, the height of ball 1 is
h1=20×3−

1

2

×10(3)2=15m