Question

31. A juggler maintains 10 balls in motion, making
each of them to rise at a height of 80 m from his
hands. To keep proper distance between them, the
time interval maintained by juggler would be
(1) 0.6 s
(2) 0.8 s
(3) 1.0 s
(4) 1.2 s​

Answer 1

Answer:

❤️❤️0.8 s

So, the initial speed of the ball as it leaves the jugglers hand is 10m/s

u = 10m/s

v = 0 (Since the ball will reach 0 velocity, once it reaches the maximum height possible and then will fall back again)

a = -g = -10m/s^2.

So, the time taken by ball to reach its maximum height

v = u+at

0 = 10-10t

t = 1 sec.

A)

V²=u²+2as. V= 0 - at the top = -u²=2×10×80. g=a=10m/s =U²=1600 U=40m/s V=u+at 0=40-10t. [ acceleration downwards is -ve] t=4s t=time of ascent Total time =8s Total ball =10 Interval kept between the ball will be 8/10=0.8s❤️❤️

Explanation:

❤️❤️hello dear here is your answer mark it brainlist ok and follow me... :) ☺️☺️❤️❤️

Answer 2

31. A juggler maintains 10 balls in motion, making

each of them to rise at a height of 80 m from his

hands. To keep proper distance between them, the

time interval maintained by juggler would be

(1) 0.6 s

(2) 0.8 s

(3) 1.0 s

(4) 1.2 s