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A-balloon-is-ascending-vertically-with-an-acceleration-of-0-2-ms-2-Two-stones-are-dropped-from-it-at-an-interval-of-2-s-The-distance-between-them-when-the-second-stone-dropped-is-take-g-9-8-m




Question Number 19774 by Tinkutara last updated on 15/Aug/17
A balloon is ascending vertically with  an acceleration of 0.2 ms^(−2) . Two stones  are dropped from it at an interval of 2 s.  The distance between them when the  second stone dropped is (take g = 9.8  ms^(−2) )
$$\mathrm{A}\:\mathrm{balloon}\:\mathrm{is}\:\mathrm{ascending}\:\mathrm{vertically}\:\mathrm{with} \\ $$$$\mathrm{an}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{0}.\mathrm{2}\:\mathrm{ms}^{−\mathrm{2}} .\:\mathrm{Two}\:\mathrm{stones} \\ $$$$\mathrm{are}\:\mathrm{dropped}\:\mathrm{from}\:\mathrm{it}\:\mathrm{at}\:\mathrm{an}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{2}\:\mathrm{s}. \\ $$$$\mathrm{The}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{them}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{stone}\:\mathrm{dropped}\:\mathrm{is}\:\left(\mathrm{take}\:{g}\:=\:\mathrm{9}.\mathrm{8}\right. \\ $$$$\left.\mathrm{ms}^{−\mathrm{2}} \right) \\ $$
Answered by ajfour last updated on 15/Aug/17
For first stone:  y_1 =h+ut−(1/2)gt^2       where h is height of balloon at t=0  u is initial velocity of balloon at t=0  For second stone:  y_2 =ut+(1/2)at^2   so y_2 −y_1 =(1/2)(g+a)t^2      y_(rel) =((1/2)×10×4)m =20m .
$$\mathrm{For}\:\mathrm{first}\:\mathrm{stone}: \\ $$$$\mathrm{y}_{\mathrm{1}} =\mathrm{h}+\mathrm{ut}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \:\:\:\: \\ $$$$\mathrm{where}\:\mathrm{h}\:\mathrm{is}\:\mathrm{height}\:\mathrm{of}\:\mathrm{balloon}\:\mathrm{at}\:\mathrm{t}=\mathrm{0} \\ $$$$\mathrm{u}\:\mathrm{is}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{balloon}\:\mathrm{at}\:\mathrm{t}=\mathrm{0} \\ $$$$\mathrm{For}\:\mathrm{second}\:\mathrm{stone}: \\ $$$$\mathrm{y}_{\mathrm{2}} =\mathrm{ut}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{at}^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{y}_{\mathrm{2}} −\mathrm{y}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{g}+\mathrm{a}\right)\mathrm{t}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{y}_{\mathrm{rel}} =\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{4}\right)\mathrm{m}\:=\mathrm{20m}\:. \\ $$
Commented by Tinkutara last updated on 15/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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