Question

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

$15,20,15$

Time taken by the small ball to return to the hands of the juggler is $\frac{2v}{g}=\frac{2\times 20}{10}=4s$. So, he is throwing the balls after $1s$ each. Let at some instant, he throws the ball number $4$. Before $1s$ of throwing it, he throws ball $3$. So, the height of ball $3$ is

$h=ut-\frac{1}{2}g{t}^2$

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

Before $2s$, he throws ball $2$. So, the height of ball $2$ is

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

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

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

$\Rightarrow h_1=15m$