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

Answer 1

Answer: (B)

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