Question

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

• 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 $=\frac{2u}{g}=\frac{2\times 20}{10}=4s$.

So he is throwing the balls after each $1s$.

Let at some instant he is throwing ball number $4$.

Before $1s$ of it he throws ball $3$. So height of ball $3$:

$h_3=20\times 1-\frac{1}{2}10(1{)}^2=15m$

Before $2s$, he throws ball $2$.

So height of ball $2$:

$h_2=20\times 2-\frac{1}{2}10(2{)}^2=20m$

Before $3s$, he throws ball $1$. So height of ball $1$:

$h_1=20\times 3-\frac{1}{2}10(3{)}^2=15m$