Question

A juggler throws 4 balls each second such that one
ball is thrown whenever the previous one is at its
highest point. The maximum height (in m) upto
which balls would rise is (where g is acceleration
due to gravity in SI units)
(1) g/32
(2) g/16
(3) g/8
(4) g/2​

Answer 1

Time after which juggler throws ball (t)= 1s
The ball reaches at it’s maximum height in 1s as he throws ball after 1s.
Maximum height is attained when final velocity becomes zero.
Let v be the final velocity and u be the initial velocity.
v=0
Acceleration is given to be g. Negative sign(-) would be used to indicate it’s in opposite direction of motion
Using first equation of motion:-
v=u+at
0=u-g
g=u
Using second equation of motion:-
s=ut+1/2at^2
s=g-1/2g=g/2
Therefore maximum height is g/2.