A juggler keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand (speed$$ $$=$$ $$20$$ $$m/s), the positions of other balls will be (take$$ g $$=$$ $$10$$ $$m/s2)

Question

Infinity Learn · Accepted Answer

The time taken by the ball to reach its maximum height is given $$byv=u-gt$$ or $$0$$ $$=20$$ $$-10$$ x t∴ $$t=2sThe$$ positions of different balls are shown in figureThe maximum distance travelled in given by $$h=ut$$−$$12gt2or$$ $$h=20$$×$$2$$−$$12$$×$$10$$×$$(2)2=20mDistance$$ travelled by first $$ballh1=12gt2=12$$×$$10$$×$$(1)2=5mh2=12gt2=12$$×$$10$$×$$(2)2=20mThe$$ distances from ground a.$$re15m$$,$$20m$$,$$15m$$

A juggler keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand (speed = 20 m/s), the positions of other balls will be (take g = 10 m/s²)

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A juggler keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand (speed = 20 m/s), the positions of other balls will be (take g = 10 m/s2)

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A juggler keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand (speed = 20 m/s), the positions of other balls will be (take g = 10 m/s²)