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

Answer 1

Answer: (B)

Time taken by same ball to return to the hands of juggler
$=\frac{2 \mu}{g}=\frac{2 \times 20}{10}=4 s$.
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 $1$ :
$h_{1}=20 \times 3-\frac{1}{2} 10(3)^{2}=15\, m$