Question

A juggler maintains four balls in motion making each of them to rise of height of $40 \,m$ 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 $V =0$
$V^{2}-u^{2}=2 a s$
$0-u^{2}=2 \times-g \times 40$
$-u^{2}=-80 \times 10$
$u =\sqrt{800}$
$u =2 \sqrt{2} \times 10=20 \sqrt{2}$
Displacement of the ball is zero.
So, $S = ut +\frac{1}{2}(- g ) t ^{2}$
(g is negative because acts in opposite direction)
$0=20 \sqrt{2} t -\frac{1}{2} \times 10 t ^{2}$
$20 \sqrt{2} t =5 t ^{2}$
$\frac{20 \sqrt{2}}{5}=t$
$4 \sqrt{2}=t$
$\therefore $ Time interval of each ball
$\frac{4 \sqrt{2}}{4}=\sqrt{2}=1.414 \sec$