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Question Number 94602 by peter frank last updated on 19/May/20

Answered by mr W last updated on 21/May/20

Commented by mr W last updated on 21/May/20

$$attimet:$$$$x_M=d$$$$y_M=h-\frac{1}{2}{gt}^2$$$$x_B=u\cos \theta t$$$$y_B=u\sin \theta t-\frac{1}{2}{gt}^2$$$$suchthatthebulletmeetsthemonkey,$$$$u\cos \theta t=d$$$$\Rightarrow t=\frac{d}{u\cos \theta }$$$$u\sin \theta t-\frac{1}{2}{gt}^2=h-\frac{1}{2}{gt}^2$$$$u\sin \theta \times \frac{d}{u\cos \theta }=h$$$$\Rightarrow \tan \theta =\frac{h}{d}$$$$thisistrueifthegunisdirectedto$$$$themonkeyasitfiresthebullet.$$$$thatmeansthe\mathit{b}\mathit{u}\mathit{l}\mathit{l}\mathit{e}\mathit{t}\mathit{c}\mathit{a}\mathit{n}\mathit{a}\mathit{l}\mathit{w}\mathit{a}\mathit{y}\mathit{s}$$$$\mathit{m}\mathit{e}\mathit{e}\mathit{t}\mathit{t}\mathit{h}\mathit{e}\mathit{m}\mathit{o}\mathit{n}\mathit{k}\mathit{e}\mathit{y}.$$$$$$$$suchthatthebulletmeetsthemonkey$$$$intheair:$$$$h-\frac{1}{2}g{\left(\frac{d}{u\cos \theta }\right)}^2⩾0$$$$u^2⩾\frac{{gd}^2}{2h}(1+{\tan }^2\theta )=\frac{{gd}^2}{2h}(1+\frac{h^2}{d^2})=\frac{g(d^2+h^2)}{2h}$$$$\Rightarrow u⩾\sqrt{\frac{g(d^2+h^2)}{2h}}$$

Commented by peter frank last updated on 22/May/20

$$\mathrm{thank}\mathrm{you}$$

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