Question

A juggler keeps moving four balls in air throwing the ball after regular intervals. when one leaves his hand (speed =40m/s ), find the position of other balls (height in meter ) . (take g =10 m/s^2)

Answer 1

Answer:

Time taken by the small ball to return to the hands of the juggler is g2v=102×20=4s. So, he is throwing the balls after 1s each. Let at some instant, he throws the ball number 4. Before 1s of throwing it, he throws ball 3. So, the height of ball 3 is

h=ut−21gt2

h3=20×1−21(10)(1)2=15m

Before 2s, he throws ball 2. So, the height of ball 2 is

h2=20×2−21×10(2)2=20m

Before 3s, he throws ball 1. So, the height of ball 1 is

h1=20×3−21×10(3)2

⇒h1=15m