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Question Number 27044 by Tinkutara last updated on 01/Jan/18

Answered by mrW1 last updated on 02/Jan/18

$$\alpha =30°$$$$x=u\cos \theta t$$$$y=u\sin \theta t-\frac{{gt}^2}{2}$$$$bylanding:$$$$y=x\tan \alpha =\frac{x\sin \alpha }{\cos \alpha }$$$$t_1=3.5s$$$$\Rightarrow {ut}_1\sin \theta -\frac{{gt}_1^2}{2}=\frac{\sin \alpha }{\cos \alpha }\times {ut}_1\cos \theta$$$$\Rightarrow \sin \theta -\frac{{gt}_1}{2u}=\frac{\sin \alpha }{\cos \alpha }\cos \theta$$$$\Rightarrow \cos \alpha \sin \theta -\frac{{gt}_1\cos \alpha }{2u}=\sin \alpha \cos \theta$$$$\Rightarrow \cos \alpha \sin \theta -\sin \alpha \cos \theta =\frac{{gt}_1\cos \alpha }{2u}$$$$\Rightarrow \sin (\theta -\alpha )=\frac{{gt}_1\cos \alpha }{2u}$$$$\Rightarrow \theta =\alpha +{\sin }^{-1}\left(\frac{{gt}_1\cos \alpha }{2u}\right)$$$$\Rightarrow \theta =30+{\sin }^{-1}\left(\frac{10\times 3.5\times \cos 30°}{2\times 53}\right)=46.62°$$$$$$$$att=2s:$$$$u_x\left(t\right)=u\cos \theta$$$$u_y\left(t\right)=u\sin \theta -gt$$$$u\left(t\right)=\sqrt{{\left(u\cos \theta \right)}^2+{(u\sin \theta -gt)}^2}$$$$u\left(t\right)=\sqrt{u^2+gt(gt-2u\sin \theta )}$$$$u\left(t\right)=\sqrt{{53}^2+10\times 2(10\times 2-2\times 53\times \sin 46.62°)}=40.84m/s$$$$\mathrm{\Delta }u=53-40.84=12.16m/s$$

Commented by mrW1 last updated on 03/Jan/18

$$Itcanalsobeunderstoodlikethis:$$$$\mathrm{\Delta }{u}_x=0$$$$\mathrm{\Delta }{u}_y=-gt=-10\times 2=-20m/s$$$$\Rightarrow \mid \mathrm{\Delta }u\mid =\sqrt{0^2+{(-20)}^2}=20m/s$$

Commented by Tinkutara last updated on 03/Jan/18

Yes answer given is 20 m/s. Thank you very much Sir! I got the answer.

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