Question

A juggler keeps moving four balls in air throwing the ball after regular intervals. when one leaves his hand (speed =20m/s ), find the kinetic energy of first ball when fourth ball is in his hand (take g =10 m/s^2)

Answer 1

Explanation:

Initial speed when ball leaves his hand,

u = 20m/s

v = 0

a = -g = -10m/s²

time taken to reach maximum height

v = u + at

0 = 20-10t

t = 2 sec

so total time for which the ball will be in air is 4 sec

since the juggler is juggling 4 balls at one time, each ball is spaced at 1 sec interval.

since the final velocity is zero, therefore the kinetic energy of the ball will be zero.