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

$$\mathrm{From}\mathrm{a}\mathrm{tower}\mathrm{of}\mathrm{height}H,\mathrm{a}\mathrm{particle}\mathrm{is}$$$$\mathrm{thrown}\mathrm{vertically}\mathrm{upward}\mathrm{with}\mathrm{speed}$$$$u.\mathrm{The}\mathrm{time}\mathrm{taken}\mathrm{by}\mathrm{the}\mathrm{particle},\mathrm{to}$$$$\mathrm{hit}\mathrm{the}\mathrm{ground},\mathrm{is}n\mathrm{times}\mathrm{that}\mathrm{taken}\mathrm{by}$$$$\mathrm{it}\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{highest}\mathrm{point}\mathrm{of}\mathrm{its}\mathrm{path}.$$$$\mathrm{The}\mathrm{relation}\mathrm{between}H,u\mathrm{and}n\mathrm{is}$$$$\left(1\right)2gH=n^2{u}^2$$$$\left(2\right)gH={(n-2)}^2{u}^2$$$$\left(3\right)2gH={nu}^2(n-2)$$$$\left(4\right)gH=(n-2)u^2$$

Answered by ajfour last updated on 15/Jul/17

$$-\mathrm{H}=\mathrm{u}\left(\mathrm{nt}\right)-\frac{1}{2}\mathrm{g}{\left(\mathrm{nt}\right)}^2\dots .\left(\mathrm{i}\right)$$$$\mathrm{Also}0=\mathrm{u}-\mathrm{gt}\Rightarrow \mathrm{t}=\mathrm{u}/\mathrm{g}$$$$\mathrm{substituting}\mathrm{in}\left(\mathrm{i}\right):$$$$-\mathrm{H}=\frac{{\mathrm{nu}}^2}{\mathrm{g}}-\frac{{\mathrm{n}}^2{\mathrm{u}}^2}{2\mathrm{g}}\mathrm{or}$$$${\mathrm{n}}^2{\mathrm{u}}^2-{2\mathrm{nu}}^2=2\mathrm{gH}\mathrm{or}$$$$2\mathrm{gH}={\mathrm{nu}}^2(\mathrm{n}-2)$$$$\left[\mathrm{option}\left(3\right)\right].$$

Commented by Tinkutara last updated on 15/Jul/17

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

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