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

Question

A juggler keeps on moving four balls in air throwing the balls after regular intervals. When one ball leaves his hand (speed =20 ms -1 ), the position of other balls (height in metre) 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 the small ball to return to the hands of the juggler is

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

.
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 = u t - \frac{1}{2} g t^{2}

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

Before

2 s

, he throws ball

2

.
So, the height of ball

2

is

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

Before

3 s

, he throws ball

1

.
So, the height of ball

1

is

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

\Rightarrow h_{1} = 15 m

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