Question

A juggler maintains four balls in motion making each of them to rise of height of $40m$ from his hand, what time interval should be maintain for the proper distance between them?

• A. 1.71 seconds
• B. 2.14 seconds
• C. 1.41 seconds
• D. 4 seconds

Answer 1

Answer: (C)

If $t$ is the total time of flight of ball in going up and
coming back, then total displacement in time $t$ is zero
because ball comes back in hand of juggler.
When ball is going at the highest point then $V=0$
$V^2-u^2=2as$
$0-u^2=2\times -g\times 40$
$-u^2=-80\times 10$
$u=\sqrt{800}$
$u=2\sqrt{2}\times 10=20\sqrt{2}$
Displacement of the ball is zero.
So, $S=ut+\frac{1}{2}(-g)t^2$
(g is negative because acts in opposite direction)
$0=20\sqrt{2}t-\frac{1}{2}\times 10t^2$
$20\sqrt{2}t=5t^2$
$\frac{20\sqrt{2}}{5}=t$
$4\sqrt{2}=t$
$\therefore$ Time interval of each ball
$\frac{4\sqrt{2}}{4}=\sqrt{2}=1.414\sec$