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From-a-tower-of-height-H-a-particle-is-thrown-vertically-upward-with-speed-u-The-time-taken-by-the-particle-to-hit-the-ground-is-n-times-that-taken-by-it-to-reach-the-highest-point-of-its-path-Th

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Question Number 18140 by Tinkutara last updated on 15/Jul/17
From a tower of height H, a particle is thrown vertically upward with speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (1) 2gH = n^2 u^2 (2) gH = (n − 2)^2 u^2 (3) 2gH = nu^2 (n − 2) (4) gH = (n − 2)u^2
$$\mathrm{From}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{of}\:\mathrm{height}\:{H},\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{speed} \\ $$$${u}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle},\:\mathrm{to} \\ $$$$\mathrm{hit}\:\mathrm{the}\:\mathrm{ground},\:\mathrm{is}\:{n}\:\mathrm{times}\:\mathrm{that}\:\mathrm{taken}\:\mathrm{by} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{highest}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}. \\ $$$$\mathrm{The}\:\mathrm{relation}\:\mathrm{between}\:{H},\:{u}\:\mathrm{and}\:{n}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{gH}\:=\:{n}^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right)^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}{gH}\:=\:{nu}^{\mathrm{2}} \left({n}\:−\:\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right){u}^{\mathrm{2}} \\ $$
Answered by ajfour last updated on 15/Jul/17
−H=u(nt)−(1/2)g(nt)^2 ....(i) Also 0=u−gt ⇒ t=u/g substituting in (i): −H=((nu^2 )/g)−((n^2 u^2 )/(2g)) or n^2 u^2 −2nu^2 =2gH or 2gH=nu^2 (n−2) [option (3)] .
$$−\mathrm{H}=\mathrm{u}\left(\mathrm{nt}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{nt}\right)^{\mathrm{2}} \:\:\:\:….\left(\mathrm{i}\right) \\ $$$$\mathrm{Also}\:\:\:\:\mathrm{0}=\mathrm{u}−\mathrm{gt}\:\:\:\Rightarrow\:\:\:\mathrm{t}=\mathrm{u}/\mathrm{g} \\ $$$$\mathrm{substituting}\:\mathrm{in}\:\left(\mathrm{i}\right): \\ $$$$−\mathrm{H}=\frac{\mathrm{nu}^{\mathrm{2}} }{\mathrm{g}}−\frac{\mathrm{n}^{\mathrm{2}} \mathrm{u}^{\mathrm{2}} }{\mathrm{2g}}\:\:\mathrm{or} \\ $$$$\mathrm{n}^{\mathrm{2}} \mathrm{u}^{\mathrm{2}} −\mathrm{2nu}^{\mathrm{2}} =\mathrm{2gH}\:\:\mathrm{or} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2gH}=\mathrm{nu}^{\mathrm{2}} \left(\mathrm{n}−\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{option}\:\:\left(\mathrm{3}\right)\right]\:\:. \\ $$
Commented by Tinkutara last updated on 15/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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