Question

A juggler keeps on moving four balls in air by throwing the balls after regular intervals. He throws the ball with a speed of 40 m/s. When one ball leaves his hand, the positions of other balls from his hand will be (in m)

Answer 1

Step-by-step explanation:

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