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Question Number 20915 by Tinkutara last updated on 07/Sep/17

$$\mathrm{Two}\mathrm{shells}\mathrm{are}\mathrm{fired}\mathrm{from}\mathrm{a}\mathrm{canon}\mathrm{with}\mathrm{speed}\mathrm{u}\mathrm{each},\mathrm{at}$$$$\mathrm{angles}\mathrm{of}\alpha \mathrm{and}\beta \mathrm{respectively}\mathrm{with}\mathrm{the}\mathrm{horizontal}.\mathrm{The}$$$$\mathrm{time}\mathrm{interval}\mathrm{between}\mathrm{the}\mathrm{shots}\mathrm{is}t.\mathrm{They}\mathrm{collide}\mathrm{in}\mathrm{mid}$$$$\mathrm{air}\mathrm{after}\mathrm{time}T\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{shot}.\mathrm{Which}\mathrm{of}\mathrm{the}$$$$\mathrm{following}\mathrm{conditions}\mathrm{must}\mathrm{be}\mathrm{satisfied}?$$$$\left(a\right)\alpha >\beta$$$$\left(b\right)T\cos \alpha =(T-t)\cos \beta$$$$\left(c\right)(T-t)\cos \alpha =T\cos \beta$$$$\left(d\right)u\sin \alpha T-\frac{1}{2}g{T}^2=u\sin \beta (T-t)-\frac{1}{2}g{(T-t)}^2$$

Answered by ajfour last updated on 08/Sep/17

$$AttimeT{x}_{\alpha }=x_{\beta }$$$$\left(u\cos \beta \right)(T-t)=\left(u\cos \alpha \right)T$$$$\Rightarrow T\cos \alpha =(T-t)\cos \beta$$$$\mathit{h}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\left(\mathit{a}\right)needbesatisfied$$$$alsoattimeT,y_{\alpha }=y_{\beta }$$$$\mathit{h}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\left(\mathit{d}\right)shouldalsobe$$$$satisfied.$$

Commented by Tinkutara last updated on 08/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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