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)
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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
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