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Question Number 91946 by Rio Michael last updated on 03/May/20

$$\mathrm{a}\mathrm{particle}\mathrm{is}\mathrm{projected}\mathrm{from}\mathrm{a}\mathrm{point}\mathrm{at}\mathrm{a}\mathrm{height}3h\mathrm{metres}\mathrm{above}\mathrm{a}\mathrm{horizontal}$$$$\mathrm{play}\mathrm{ground}.\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{the}\mathrm{projectile}\mathrm{makes}\mathrm{an}\mathrm{angle}\alpha \mathrm{with}\mathrm{the}$$$$\mathrm{horizontal}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{projection}.\mathrm{show}\mathrm{that}\mathrm{if}\mathrm{th}\mathrm{greatest}$$$$\mathrm{height}\mathrm{reached}\mathrm{above}\mathrm{the}\mathrm{point}\mathrm{lc}\mathrm{projection}\mathrm{is}h\mathrm{metres},\mathrm{then}\mathrm{the}\mathrm{horizontal}$$$$\mathrm{distance}\mathrm{travelled}\mathrm{by}\mathrm{the}\mathrm{particle}\mathrm{before}\mathrm{striking}\mathrm{the}\mathrm{plane}\mathrm{is}6h\cot \alpha \mathrm{metres}.$$$$\mathrm{Find}\mathrm{the}\mathrm{vertical}\mathrm{and}\mathrm{horizontal}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{just}$$$$\mathrm{before}\mathrm{it}\mathrm{hits}\mathrm{the}\mathrm{ground}.$$

Commented by mr W last updated on 04/May/20

$$applymethodsasinQ91655$$

Commented by Rio Michael last updated on 04/May/20

$$\mathrm{oh}\mathrm{sir}\mathrm{am}\mathrm{sorry}\mathrm{i}\mathrm{will}\mathrm{fix}\mathrm{that}\mathrm{next}\mathrm{time}.\mathrm{and}\mathrm{thanks}$$

Commented by Rio Michael last updated on 04/May/20

$$\mathrm{sir}\mathrm{in}\mathrm{this}\mathrm{question},\mathrm{should}\mathrm{i}\mathrm{assume}$$$$\mathrm{the}\mathrm{initial}\mathrm{velocity}\mathrm{is}u,\mathrm{and}\mathrm{apply}\mathrm{that}\mathrm{method}?$$$$\mathrm{It}\mathrm{seems}\mathrm{it}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{hiegt}\mathrm{reason}\mathrm{why}\mathrm{applying}\mathrm{the}$$$$\mathrm{parabolic}\mathrm{property}\mathrm{method}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{work}.$$

Answered by mr W last updated on 04/May/20

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}$$

Commented by Rio Michael last updated on 04/May/20

$$\mathrm{sir}\mathrm{i}\mathrm{have}\mathrm{some}\mathrm{questions}\mathrm{about}\mathrm{this}\mathrm{method}\mathrm{of}\mathrm{yours}.$$$$-\mathrm{do}\mathrm{you}\mathrm{assume}\mathrm{that}\mathrm{the}\mathrm{origin}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{vertex}$$$$\mathrm{that}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{point}h\mathrm{on}\mathrm{your}\mathrm{diagram}?\mathrm{or}\mathrm{at}\mathrm{O}?$$$$-\mathrm{also}\mathrm{when}\mathrm{using}\mathrm{your}\mathrm{expression}\frac{y_1}{y_2}={\left(\frac{x_1}{x_2}\right)}^2\mathrm{how}\mathrm{do}$$$$\mathrm{you}\mathrm{choose}\mathrm{your}\mathrm{points}?\mathrm{that}\mathrm{is}\mathrm{your}\mathrm{coordinates}\left(x_1{y}_1\right)\mathrm{and}$$$$(x_2,y_2)?$$

Commented by mr W last updated on 04/May/20

Commented by mr W last updated on 04/May/20

$$thepathoftheparticleisABCD.$$$$Bisthehighestpoint.theparabola$$$$issymmetricaboutB.wecanimage$$$$theparticlestartsfromO.OAis$$$$themirrorimageofCD.$$$$\tan \alpha =\frac{2h}{\frac{AC}{2}}$$$$\Rightarrow \frac{AC}{2}=2h\cot \alpha$$$$\frac{3h+h}{h}={\left(\frac{{OB}^{\prime }}{A^{\prime }{B}^{\prime }}\right)}^2=4$$$${OB}^{\prime }=2A^{\prime }{B}^{\prime }=AC=4h\cot \alpha =B^{\prime }D$$$$A^{\prime }D=A^{\prime }{B}^{\prime }+B^{\prime }D=\frac{AC}{2}+B^{\prime }D$$$$A^{\prime }D=2h\cot \alpha +4h\cot \alpha =6h\cot \alpha$$$$\Rightarrow proved$$$$$$$$\tan \beta =\frac{2(3h+h)}{B^{\prime }D}=\frac{8h}{4h\cot \alpha }=2\tan \alpha$$$$u_{Dy}=\sqrt{2g(h+3h)}=2\sqrt{2gh}$$$$u_{Dx}=\frac{v_{Dx}}{\tan \beta }=\frac{2\sqrt{2gh}}{2\tan \alpha }=\cot \alpha \sqrt{2gh}$$

Commented by mr W last updated on 04/May/20

Commented by Rio Michael last updated on 04/May/20

$$\mathrm{thanks}\mathrm{again}\mathrm{sir}$$

Commented by Tawa11 last updated on 06/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 04/May/20

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}\mathit{I}$$

Commented by mr W last updated on 04/May/20

Commented by mr W last updated on 04/May/20

$$x=u\cos \alpha t$$$$y=u\sin \alpha t-\frac{1}{2}{gt}^2$$$$max.hightatt=t_1:$$$$u\sin \alpha -{gt}_1=0\Rightarrow t_1=\frac{u\sin \alpha }{g}$$$$y_B=u\sin \alpha t_1-\frac{1}{2}{gt}_1^2=h$$$$(u\sin \alpha -\frac{1}{2}g\times \frac{u\sin \alpha }{g})\frac{u\sin \alpha }{g}=h$$$$\Rightarrow u\sin \alpha =\sqrt{2gh}...\left(i\right)$$$$$$$$ballhitsthegroundatt=t_2:$$$$y_D=u\sin \alpha t_2-\frac{1}{2}{gt}_2^2=-3h$$$${gt}_2^2-2\sqrt{2gh}{t}_2-6h=0$$$$\Rightarrow t_2=3\sqrt{\frac{2h}{g}}$$$$x_D=u\cos \alpha t_2=3\sqrt{\frac{2h}{g}}u\cos \alpha$$$$\Rightarrow u\cos \alpha =\frac{x_D}{3\sqrt{\frac{2h}{g}}}...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\tan \alpha =\frac{6h}{x_D}$$$$\Rightarrow x_D=6h\cot \alpha \Rightarrow proved$$$$$$$$u_{Dx}=u\cos \alpha =\frac{6h\cot \alpha }{3\sqrt{\frac{2h}{g}}}=\cot \alpha \sqrt{2gh}$$$$u_{Dy}=u\sin \alpha -{gt}_2=\sqrt{2gh}-3\sqrt{2gh}=-2\sqrt{2gh}$$

Commented by mr W last updated on 04/May/20

$$thesameresultaswithmethodI.$$

Commented by Rio Michael last updated on 04/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}$$

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