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Question Number 135653 by I want to learn more last updated on 14/Mar/21

Commented by mr W last updated on 14/Mar/21

$$questioniswrong.justcheckagain.$$

Commented by I want to learn more last updated on 14/Mar/21

$$\mathrm{That}\mathrm{is}\mathrm{the}\mathrm{question}\mathrm{sir},\mathrm{and}\mathrm{what}\mathrm{makes}\mathrm{it}\mathrm{wrong}\mathrm{sir}\mathrm{please}.$$

Commented by mr W last updated on 14/Mar/21

$$just\underset{―}{\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}}:canyoustand2.5maway$$$$fromabuildingandthrowastone$$$$atanangle60°andhitapointonthe$$$$building16maboveyou?$$

Commented by mr W last updated on 14/Mar/21

Commented by I want to learn more last updated on 14/Mar/21

$$\mathrm{Alright}\mathrm{sir},\mathrm{i}\mathrm{understand}.$$

Commented by I want to learn more last updated on 14/Mar/21

$$\mathrm{I}\mathrm{get}\mathrm{it}\mathrm{clearly}\mathrm{sir}.$$

Commented by I want to learn more last updated on 14/Mar/21

$$\mathrm{Sir}\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{correct}\mathrm{the}\mathrm{value},\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{see}\mathrm{the}\mathrm{procedure}$$$$\mathrm{and}\mathrm{diagram}\mathrm{for}\mathrm{such}\mathrm{question}$$

Commented by mr W last updated on 14/Mar/21

$$soiaskedyoutocheckagain!$$

Commented by I want to learn more last updated on 14/Mar/21

$$\mathrm{It}\mathrm{will}\mathrm{help}\mathrm{me}\mathrm{solve}\mathrm{relatively}\mathrm{questions}\mathrm{here}.$$

Answered by mr W last updated on 15/Mar/21

Commented by mr W last updated on 15/Mar/21

$$timefromAtoBist$$$$t=\frac{L}{u\cos \theta }$$$$h=u\sin \theta t-\frac{1}{2}{gt}^2$$$$h=u\sin \theta \times \frac{L}{u\cos \theta }-\frac{1}{2}g{\left(\frac{L}{u\cos \theta }\right)}^2$$$$h=L\tan \theta -\frac{{gL}^2}{2u^2}\times \frac{1}{{\cos }^2\theta }$$$$\frac{h}{L}=\tan \theta -\frac{1}{2}\times \frac{gL}{u^2{\cos }^2\theta }$$$$\frac{h}{L}=\tan \theta -\frac{gL}{2u^2}(1+{\tan }^2\theta )$$$$\Rightarrow u=\sqrt{\frac{gL(1+{\tan }^2\theta )}{2(\tan \theta -\frac{h}{L})}}$$$$withh=16m,L=25m,\theta =60°weget$$$$u=\sqrt{\frac{10\times 25\times (1+{\left(\sqrt{3}\right)}^2)}{2(\sqrt{3}-\frac{16}{25})}}\approx 21.397m/s$$$$t=\frac{25}{21.397\times \cos 60°}=2.337s$$$$$$$$velocityatpointBis{v}_B$$$$v_{B,x}=u\cos \theta$$$$v_{B,y}=u\sin \theta -gt$$$$v_B=\sqrt{{\left(u\cos \theta \right)}^2+{(u\sin \theta -gt)}^2}$$$$=\sqrt{u^2-2g(u\sin \theta t-\frac{1}{2}{gt}^2)}$$$$=\sqrt{u^2-2gh}$$$$=\sqrt{21.{397}^2-2\times 10\times 16}\approx 11.74m/s$$$$$$$$say\beta =angleof{v}_{B}abovehorizontal$$$$\beta >0:\nearrow$$$$\beta =0:\to$$$$\beta <0:\searrow$$$$\tan \beta =\frac{v_{B,y}}{v_{B,x}}=\frac{u\sin \theta -gt}{u\cos \theta }$$$$=\tan \theta -\frac{gL}{u^2{\cos }^2\theta }$$$$=\tan \theta -2(\tan \theta -\frac{h}{L})$$$$=\frac{2h}{L}-\tan \theta$$$$\Rightarrow \beta ={\tan }^{-1}(\frac{2h}{L}-\tan \theta )$$$$={\tan }^{-1}(\frac{2\times 16}{25}-\sqrt{3})\approx -24.325°$$$$$$$$max.height{h}_{max}$$$${gh}_{max}=\frac{1}{2}{\left(u\sin \theta \right)}^2$$$$h_{max}=\frac{u^2{\sin }^2\theta }{2g}$$$$=\frac{L(1+{\tan }^2\theta ){\sin }^2\theta }{4(\tan \theta -\frac{h}{L})}$$$$=\frac{25\times (1+3){\left(\frac{\sqrt{3}}{2}\right)}^2}{4(\sqrt{3}-\frac{16}{25})}\approx 17.169m>h$$$$thatmeansmax.heightisreached$$$$beforestrikingthewall.$$

Commented by mr W last updated on 15/Mar/21

Commented by I want to learn more last updated on 15/Mar/21

$$\mathrm{Wow},\mathrm{i}\mathrm{really}\mathrm{understand}\mathrm{sir}.\mathrm{I}\mathrm{have}\mathrm{used}\mathrm{it}\mathrm{to}\mathrm{solve}\mathrm{questions}\mathrm{in}\mathrm{the}\mathrm{same}$$$$\mathrm{category}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Commented by Tawa11 last updated on 14/Sep/21

$$\mathrm{nice}$$

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