A-Juggler-is-maintaining-4-balls-in-motion-making-each-in-term-to-rise-to-height-of-20-m-Which-of-following-position-is-not-possible-for-the-balls-when-one-ball-is-just-leaving-his-hand-1-5-m-2-

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Question Number 15117 by Tinkutara last updated on 07/Jun/17

$$\mathrm{A}\mathrm{Juggler}\mathrm{is}\mathrm{maintaining}4\mathrm{balls}\mathrm{in}$$$$\mathrm{motion}\mathrm{making}\mathrm{each}\mathrm{in}\mathrm{term}\mathrm{to}\mathrm{rise}\mathrm{to}$$$$\mathrm{height}\mathrm{of}20\mathrm{m}.\mathrm{Which}\mathrm{of}\mathrm{following}$$$$\mathrm{position}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{the}\mathrm{balls},$$$$\mathrm{when}\mathrm{one}\mathrm{ball}\mathrm{is}\mathrm{just}\mathrm{leaving}\mathrm{his}\mathrm{hand}?$$$$\left(1\right)5\mathrm{m}$$$$\left(2\right)\mathrm{All}\mathrm{position}\mathrm{are}\mathrm{possible}$$$$\left(3\right)15\mathrm{m}$$$$\left(4\right)20\mathrm{m}$$

Answered by mrW1 last updated on 07/Jun/17

$$\mathrm{time}\mathrm{for}\mathrm{a}\mathrm{cycle}:$$$$20=\frac{1}{2}\times 10\times {\left(\frac{\mathrm{T}}{2}\right)}^2$$$$\Rightarrow \mathrm{T}=4\mathrm{s}$$$$\mathrm{that}\mathrm{means}\mathrm{the}\mathrm{balls}\mathrm{leave}\mathrm{the}\mathrm{hand}$$$$\mathrm{at}\mathrm{an}\mathrm{interval}\mathrm{from}1\mathrm{second}.$$$$$$$$\mathrm{position}\mathrm{of}\mathrm{ball}1:$$$$\mathrm{h}=0\left(\mathrm{leaving}\mathrm{the}\mathrm{hand}\right)$$$$$$$$\mathrm{position}\mathrm{of}\mathrm{ball}2:$$$$\mathrm{h}=20-\frac{1}{2}\times 10\times 1^2=15\mathrm{m}$$$$\mathrm{position}\mathrm{of}\mathrm{ball}3:$$$$\mathrm{h}=20$$$$$$$$\mathrm{position}\mathrm{of}\mathrm{ball}4:$$$$\mathrm{h}=20-\frac{1}{2}\times 10\times 1^2=15\mathrm{m}$$$$$$$$\Rightarrow \mathrm{answer}\left(1\right)$$

Commented by Tinkutara last updated on 07/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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