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Question Number 15117 by Tinkutara last updated on 07/Jun/17
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) All position are possible (3) 15 m (4) 20 m
$$\mathrm{A}\:\mathrm{Juggler}\:\mathrm{is}\:\mathrm{maintaining}\:\mathrm{4}\:\mathrm{balls}\:\mathrm{in} \\ $$$$\mathrm{motion}\:\mathrm{making}\:\mathrm{each}\:\mathrm{in}\:\mathrm{term}\:\mathrm{to}\:\mathrm{rise}\:\mathrm{to} \\ $$$$\mathrm{height}\:\mathrm{of}\:\mathrm{20}\:\mathrm{m}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{following} \\ $$$$\mathrm{position}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{for}\:\mathrm{the}\:\mathrm{balls}, \\ $$$$\mathrm{when}\:\mathrm{one}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{just}\:\mathrm{leaving}\:\mathrm{his}\:\mathrm{hand}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5}\:\mathrm{m} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{position}\:\mathrm{are}\:\mathrm{possible} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{15}\:\mathrm{m} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{20}\:\mathrm{m} \\ $$
Answered by mrW1 last updated on 07/Jun/17
time for a cycle: 20=(1/2)×10×((T/2))^2 ⇒T=4 s that means the balls leave the hand at an interval from 1 second. position of ball 1: h=0 (leaving the hand) position of ball 2: h=20−(1/2)×10×1^2 =15 m position of ball 3: h=20 position of ball 4: h=20−(1/2)×10×1^2 =15 m ⇒answer (1)
$$\mathrm{time}\:\mathrm{for}\:\mathrm{a}\:\mathrm{cycle}: \\ $$$$\mathrm{20}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\left(\frac{\mathrm{T}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{T}=\mathrm{4}\:\mathrm{s} \\ $$$$\mathrm{that}\:\mathrm{means}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{leave}\:\mathrm{the}\:\mathrm{hand} \\ $$$$\mathrm{at}\:\mathrm{an}\:\mathrm{interval}\:\mathrm{from}\:\mathrm{1}\:\mathrm{second}. \\ $$$$ \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{ball}\:\mathrm{1}: \\ $$$$\mathrm{h}=\mathrm{0}\:\left(\mathrm{leaving}\:\mathrm{the}\:\mathrm{hand}\right) \\ $$$$ \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{ball}\:\mathrm{2}: \\ $$$$\mathrm{h}=\mathrm{20}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{1}^{\mathrm{2}} =\mathrm{15}\:\mathrm{m} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{ball}\:\mathrm{3}: \\ $$$$\mathrm{h}=\mathrm{20} \\ $$$$ \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{ball}\:\mathrm{4}: \\ $$$$\mathrm{h}=\mathrm{20}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{1}^{\mathrm{2}} =\mathrm{15}\:\mathrm{m} \\ $$$$ \\ $$$$\Rightarrow\mathrm{answer}\:\left(\mathrm{1}\right) \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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