A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed = 20 ms-1) the position of other balls (height in m) will be (Take g = 10 ms-2)

Question

A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed = 20 ms-1) the position of other balls (height in m) will be (Take g = 10 ms-2) (A) 10,20,10 (B) 15,20,15 (C) 5,15,20 (D) 5,10,20

Tardigrade Admin · Accepted Answer

Time taken by same ball to return to the hands of juggler

= \frac{2 μ}{g} = \frac{2 \times 20}{10} = 4 s

.
So he is throwing the balls after each

1 s

.
Let at some instant he is throwing ball number

4

.
Before

1 s

of it he throws ball. So height of ball

3

:

h_{3} = 20 \times 1 - \frac{1}{2} 10 (1)^{2} = 15 m

Before

2 s

, he throws ball

2

.
So height of ball

2

:

h_{2} = 20 \times 2 - \frac{1}{2} 10 (2)^{2} = 20 m

Before

3 s

, he throws ball

1

.
So height of ball

1

:

h_{1} = 20 \times 3 - \frac{1}{2} 10 (3)^{2} = 15 m

Q. A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed =20ms−1) the position of other balls (height in m) will be (Take g=10ms−2)

10,20,10

15,20,15

5,15,20

5,10,20

Q. A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed $= 20 m s^{- 1})$ the position of other balls (height in $m$ ) will be (Take $g = 10 m s^{- 2})$