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Question Number 55191 by Tawa1 last updated on 19/Feb/19

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

$$18)atgreatestheughtverticalcomponentof$$$$velocityiszero$$$$0^2={\left(usin\theta \right)}^2-2{gh}_1$$$$0^2={\left\{usin(90-\theta )\right\}}^2-2{gh}_2$$$$\frac{2{gh}_1}{2{gh}_2}=\frac{u^2{sin}^2\theta }{u^2{cos}^2\theta }$$$$tan\theta =\sqrt{\frac{h_1}{h_2}}$$

Commented by Tawa1 last updated on 19/Feb/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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

$$19)Range=\left(ucos\theta \right)\times \left(timeofflight\right)$$$$timeofflight=T=2t_0$$$$0=usin\theta -{gt}_0$$$$t_0=\frac{usin\theta }{g}$$$$R=\left(ucos\theta \right)\left(\frac{2usin\theta }{g}\right)=\frac{u^{2}sin2\theta }{g}$$$$R_{max}whensin2\theta =1=sin\frac{\pi }{2}$$$$\theta =\frac{\pi }{4}$$$${sin}^{-1}\left(\frac{\sqrt{2}}{2}\right)\to {sin}^{-1}\left(\frac{1}{\sqrt{2}}\right)\to \frac{\pi }{4}$$

Commented by Tawa1 last updated on 19/Feb/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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