As per the question,
The time between each throw=0.23 sec
a)
In case of three ball, the first ball is about to reach his hand, then the time of journey of each ball =3×0.23=0.69sec
So, the ball will reach to the highest point in the time (t1)=20.69=0.345sec
at the top point the final velocity(v) =0
let the initial velocity = u
let g is the gravitational acceleration
v=u−gt1
0=u−gt1
u=gt1
So, minimum height thrown by the juggler (h)=ut1−2gt12
h=gt12−2gt12=2gt12
h=29.8×0.345×0.345=0.583m
b)
In case of five ball, the first ball is about to reach his hand, then the time of journey of each ball5×0.23=1.15sec
So, the ball will reach to the highest point in the time (t2)=21.15=0.575 sec
at the highest point, velocity (v1)=0
let the initial velocity (u1) and time g is the gravitational acceleration.
⇒v2=u2−gt2
⇒0=u2−gt2
u2=gt2
Let the minimum height reached by the ball h2=u2t2−2gt22
⇒h2=gt22−2gt22=2gt22
now substituting the value of t2
h2=29.8×0.575×0.575=1.62m
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