Question

A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand (speed $= 20 \,ms^{-1})$ the positions 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{2u}{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 $3$. 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 $1$:
$h_1 = 20 \times 3 - \frac{1}{2} 10(3)^2 = 15\,m$