Question

A juggler is throwing ball vertically up in air at regular interval of 1 second Sixth ball just leaves the hand when 1 st ball comes back to his hands, then minimum speed with which he throelws the ball is​

Answer 1

Answer:

29.4 m/s

Explanation:

When he throws the sixth ball, the first ball is about to reach his hand. Hence, the time of journey of each ball is 6 seconds.

The ball will reach the highest point in 3

seconds Hence, t = 3 seconds

Final velocity at highest point Let the initial velocity be u.

(v) = 0 m/s

Acceleration due to gravity = - 9.8 m/s^2

Using first equation of motion:

v = u + at

0= u - 9.8 x 3

u = 29.4 m/s

Hence, the maximum speed of the throw is 29.4 m/s.