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Question Number 59892 by rahul 19 last updated on 15/May/19

Commented by rahul 19 last updated on 15/May/19

$$Sir,ans.is\left(c\right)!$$

Commented by tanmay last updated on 15/May/19

$$ifoundanswer...$$

Answered by tanmay last updated on 15/May/19

Commented by tanmay last updated on 15/May/19

Commented by rahul 19 last updated on 16/May/19

$$thankssir!$$

Answered by mr W last updated on 15/May/19

Commented by mr W last updated on 15/May/19

$$u=2\sqrt{gh}$$$$fromAtoBintime{t}_1:$$$$h=u\sin \theta t_1-\frac{1}{2}{gt}_1^2...\left(i\right)$$$$a=u\cos \theta t_1...\left(ii\right)$$$$fromAtoCintime{t}_2:$$$$a+h=u\cos \theta t_2...\left(iii\right)$$$$u\sin \theta -{gt}_2=0$$$$\Rightarrow t_2=\frac{u\sin \theta }{g}$$$$\left(iii\right)-\left(ii\right):$$$$h=u\cos \theta (t_2-t_1)$$$$h=u\cos \theta (\frac{u\sin \theta }{g}-t_1)$$$$\Rightarrow t_1=\frac{u\sin \theta }{g}-\frac{h}{u\cos \theta }...\left(iv\right)$$$$put\left(iv\right)into\left(i\right):$$$$h=[u\sin \theta -\frac{1}{2}g(\frac{u\sin \theta }{g}-\frac{h}{u\cos \theta })](\frac{u\sin \theta }{g}-\frac{h}{u\cos \theta })$$$$h=\frac{1}{2g}(u^2{\sin }^2\theta -\frac{g^2{h}^2}{u^2{\cos }^2\theta })$$$$h=\frac{1}{2g}(4gh{\sin }^2\theta -\frac{g^2{h}^2}{4gh{\cos }^2\theta })$$$$1=2{\sin }^2\theta -\frac{1}{8{\cos }^2\theta }$$$$1=2-2{\cos }^2\theta -\frac{1}{8{\cos }^2\theta }$$$$16{\cos }^4\theta -8{\cos }^2\theta +1=0$$$${(4{\cos }^2\theta -1)}^2=0$$$$\Rightarrow {\cos }^2\theta =\frac{1}{4}$$$$\Rightarrow \cos \theta =\frac{1}{2}$$$$\Rightarrow \theta =\frac{\pi }{3}=60°$$$$t_2=\frac{2\sqrt{gh}}{g}\times \frac{\sqrt{3}}{2}=\sqrt{\frac{3h}{g}}$$$$t_1=\frac{2\sqrt{gh}}{g}\times \frac{\sqrt{3}}{2}-\frac{h}{2\sqrt{gh}\times \frac{1}{2}}=\sqrt{\frac{3h}{g}}-\sqrt{\frac{h}{g}}$$$$t_2-t_1=\sqrt{\frac{3h}{g}}-(\sqrt{\frac{3h}{g}}-\sqrt{\frac{h}{g}})=\sqrt{\frac{h}{g}}$$$$$$$$timefromBtoD=2(t_2-t_1)=2\sqrt{\frac{h}{g}}$$$$\Rightarrow answer\left(c\right)$$

Commented by rahul 19 last updated on 16/May/19

$$ThankYouSir!$$

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