A-stone-is-thrown-vertically-upwards-from-a-cliff-20m-high-After-a-time-of-3-s-it-passes-the-edge-of-the-cliff-on-its-way-down-Calculate-a-the-speed-of-projection-b-the-speed-when-it-hits-the-grou

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Question Number 54767 by pieroo last updated on 10/Feb/19

$$\mathrm{A}\mathrm{stone}\mathrm{is}\mathrm{thrown}\mathrm{vertically}\mathrm{upwards}$$$$\mathrm{from}\mathrm{a}\mathrm{cliff}20\mathrm{m}\mathrm{high}.\mathrm{After}\mathrm{a}\mathrm{time}\mathrm{of}3\mathrm{s}$$$$\mathrm{it}\mathrm{passes}\mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cliff}\mathrm{on}\mathrm{its}\mathrm{way}$$$$\mathrm{down}.\mathrm{Calculate}$$$$\mathrm{a})\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{projection}$$$$\mathrm{b})\mathrm{the}\mathrm{speed}\mathrm{when}\mathrm{it}\mathrm{hits}\mathrm{the}\mathrm{ground}$$$$\mathrm{c})\mathrm{the}\mathrm{times}\mathrm{when}\mathrm{it}\mathrm{is}10\mathrm{m}\mathrm{above}\mathrm{the}$$$$\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{cliff}$$$$\mathrm{d})\mathrm{the}\mathrm{time}\mathrm{it}\mathrm{is}15\mathrm{m}\mathrm{above}\mathrm{the}\mathrm{ground}$$$$(\mathrm{take}\mathrm{g}={10\mathrm{ms}}^{-2}).$$

Commented by pieroo last updated on 10/Feb/19

$$\mathrm{NEED}\mathrm{SOME}\mathrm{HELP}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

$$a)in3secondsthestonereach(maximumheightandreturn$$$$tothepointofprojection)sodisplacement=0$$$$0=ut-\frac{1}{2}{gt}^{2}sou=\frac{gt}{2}=\frac{10\times 3}{2}=15m/sec$$$$b)whenitreachattheedgeofcliffitacquire$$$$velocity=0+10\times \frac{3}{2}=15(v=u+att=\frac{3}{2}sec)$$$$requirevelocitywhenhitground$$$$v_g^2={15}^2+2\times 10\times 20=625$$$$v_g=25m/sec$$$$c)10=15\times t-\frac{1}{2}\times 10t^2[h=ut-\frac{1}{2}{gt}^2]$$$$2=3t-t^2$$$$t^2-3t+2=0\to (t-1)(t-2)=0$$$$t=1secand2sec$$$$1secwhilemovingupand2secwhenitreach$$$$sameplacemovingdown\dots$$$$d)whenitis15mabovegroundthatmeans5meter$$$$belowthebaseofcliff\dots$$$$-5=15t-\frac{1}{2}\times 10\times t^2$$$$-1=3t-t^2$$$$t^2-3t+1=0$$$$t=\frac{3+\sqrt{9-4}}{2}=\frac{3+\sqrt{5}}{2}sec$$$$plschekanswer$$$$ifnotcorrectplscomment\dots$$

Commented by pieroo last updated on 11/Feb/19

$$\mathrm{it}\mathrm{is}\mathrm{correct}.\mathrm{thanks}\mathrm{so}\mathrm{much}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Feb/19

$$mostwelcome\dots moralboost\dots$$

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