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Question Number 55418 by Tawa1 last updated on 23/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19

$$plscalculatesin2\theta =\frac{5}{13}usingcalculatr$$$$ihavenotdownloadedscientificcalcuator..$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19

$$Range=vcos\alpha \times 2t_o$$$$0=vsin\alpha -{gt}_0$$$$t_0=\frac{vsin\alpha }{g}$$$$Range=vcos\alpha \times 2\times \frac{vsin\alpha }{g}=\frac{v^{2}sin2\alpha }{g}$$$${\left(Range\right)}_{max}=\frac{v^2}{g}\times 1[maxvalueofsin2\alpha =1]$$$${\left(Range\right)}_{max}=\frac{v^2}{g}$$$$asperquestion{\left(Range\right)}_{max}=R=\frac{v^2}{g}$$

Commented by Tawa1 last updated on 24/Feb/19

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

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19

$$Range=\frac{5R}{13}=\frac{5}{13}\times \frac{v^2}{g}$$$$nowrange=\frac{v^{2}sin2\theta }{g}=\frac{5v^2}{13g}$$$$sin2\theta =\frac{5}{13}$$$$\theta =\frac{1}{2}{sin}^{-1}\left(\frac{5}{13}\right)and\pi -\frac{1}{2}{sin}^{-1}\left(\frac{5}{13}\right)$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19

$$timeofflight=2t_0=\frac{2vsin\theta }{g}$$$$$$$$[0=vsin\theta -{gt}_0]$$$$nowsin2\theta =\frac{5}{13}$$$$\theta =\frac{1}{2}{sin}^{-1}\left(\frac{5}{13}\right)and\pi -\frac{1}{2}{sin}^{-1}\left(\frac{5}{13}\right)$$$$sotimeofflight...$$$$$$$${\left(time\right)}_{flight}=\frac{2vsin\left(\frac{1}{2}{sin}^{-1}\frac{5}{13}\right)}{2g}$$$$and\frac{2vsin[\pi -\frac{1}{2}{sin}^{-1}\left(\frac{5}{13}\right)]}{2g}$$$$$$

Commented by malwaan last updated on 24/Feb/19

$${\sin }^{-1}\left(\frac{5}{13}\right)\approx 22.62$$

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Feb/19

$$thankyousir...$$

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