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Question Number 154549 by peter frank last updated on 19/Sep/21

A particle is projected with velocity 2(√(gh)) so that it just clears two walls of equal heigh(h) in the t_1 and t_(2 ) respectively.The two walls are at a distance of 2h from each other.If time passing between the two walls is 2(√(h/g)) show that (i) angle projected 60^° (ii)t_1 +t_2 =2(√((3h)/g))

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\:\mathrm{with}\:\mathrm{velocity} \\ $$$$\mathrm{2}\sqrt{\mathrm{gh}}\:\:\:\mathrm{so}\:\mathrm{that}\:\:\mathrm{it}\:\mathrm{just}\:\mathrm{clears}\:\mathrm{two} \\ $$$$\mathrm{walls}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{heigh}\left(\mathrm{h}\right)\:\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{t}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{t}_{\mathrm{2}\:} \:\mathrm{respectively}.\mathrm{The}\:\mathrm{two} \\ $$$$\mathrm{walls}\:\mathrm{are}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\:\mathrm{2h}\:\:\mathrm{from} \\ $$$$\mathrm{each}\:\mathrm{other}.\mathrm{If}\:\mathrm{time}\:\mathrm{passing}\: \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{walls}\:\mathrm{is}\:\mathrm{2}\sqrt{\frac{\mathrm{h}}{\mathrm{g}}} \\ $$$$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{i}\right)\:\mathrm{angle}\:\mathrm{projected}\:\mathrm{60}^{°} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{ii}\right)\mathrm{t}_{\mathrm{1}} +\mathrm{t}_{\mathrm{2}} =\mathrm{2}\sqrt{\frac{\mathrm{3h}}{\mathrm{g}}} \\ $$$$ \\ $$

Commented by Tawa11 last updated on 21/Sep/21

nice sir

$$\mathrm{nice}\:\mathrm{sir} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

(i)For horizontal motion between both walls; x=v_x t. But x=2h, v_x =2(√(gh))cosϑ, t=2(√(h/g)) ⇒2h=(2(√(gh))cosϑ)(2(√(h/g))) ⇒(1/2)=cosϑ⇒ϑ=60°

$$\left({i}\right)\mathrm{For}\:\mathrm{horizontal}\:\mathrm{motion}\:\mathrm{between}\:\mathrm{both}\:\mathrm{walls}; \\ $$$$\:{x}={v}_{{x}} {t}.\:\mathrm{But}\:{x}=\mathrm{2}{h},\:{v}_{{x}} =\mathrm{2}\sqrt{{gh}}\mathrm{cos}\vartheta,\:{t}=\mathrm{2}\sqrt{\frac{{h}}{{g}}} \\ $$$$\Rightarrow\mathrm{2}{h}=\left(\mathrm{2}\sqrt{{gh}}\mathrm{cos}\vartheta\right)\left(\mathrm{2}\sqrt{\frac{{h}}{{g}}}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\vartheta\Rightarrow\vartheta=\mathrm{60}° \\ $$

Commented by peter frank last updated on 20/Sep/21

thanks

$$\mathrm{thanks} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

(ii) For vertical motion: y=ut+(1/2)at^2 , y=h, u=2(√(gh))sin60°=(√(3gh)), a=−g ⇒h=(√(3gh))t−(1/2)gt^2 ⇒gt^2 −2(√(3gh))t+h=0 t=((2(√(3gh))±(√(12gh−4gh)))/(2g))=((2(√(3gh))±2(√(2gh)))/(2g)) t_1 =(√((3h)/g))+(√((2h)/g)), t_2 =(√((3h)/g))−(√((2h)/g)) t_1 +t_2 =2(√((3h)/g))

$$\left({ii}\right)\:\mathrm{For}\:\mathrm{vertical}\:\mathrm{motion}: \\ $$$${y}={ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} ,\:{y}={h},\:{u}=\mathrm{2}\sqrt{{gh}}\mathrm{sin60}°=\sqrt{\mathrm{3}{gh}},\:{a}=−{g} \\ $$$$\Rightarrow{h}=\sqrt{\mathrm{3}{gh}}{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \Rightarrow{gt}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}{gh}}{t}+{h}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{2}\sqrt{\mathrm{3}{gh}}\pm\sqrt{\mathrm{12}{gh}−\mathrm{4}{gh}}}{\mathrm{2}{g}}=\frac{\mathrm{2}\sqrt{\mathrm{3}{gh}}\pm\mathrm{2}\sqrt{\mathrm{2}{gh}}}{\mathrm{2}{g}} \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{3}{h}}{{g}}}+\sqrt{\frac{\mathrm{2}{h}}{{g}}},\:{t}_{\mathrm{2}} =\sqrt{\frac{\mathrm{3}{h}}{{g}}}−\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$${t}_{\mathrm{1}} +{t}_{\mathrm{2}} =\mathrm{2}\sqrt{\frac{\mathrm{3}{h}}{{g}}} \\ $$

Answered by peter frank last updated on 20/Sep/21

Answered by peter frank last updated on 20/Sep/21

x−direction x_1 =v_o cos θt_1 x_2 =v_o cos θt_2 v_o =2(√(gh)) x_2 −x_1 =2h (t_2 −t_1 )=2(√(h/g)) v_o cos θ(t_2 −t_1 )=2h cosθ=((2h)/(v_o (t_2 −t_1 ))) =(1/2) θ= cos^(−1) (1/2)=60^° also y−direction y=v_o sin θt_1 −(1/2)gt_1 ^2 y=v_o sin θt_2 −(1/2)gt_2 ^2 v_o sin θt_1 −(1/2)gt_1 ^2 =v_o sin θt_2 −(1/2)gt_2 ^2 2(√(gh)) sin θ(t_2 −t_1 )=(1/2)g(t_2 ^2 −t_1 ^2 ) 2(√(gh)) sin θ=(1/2)g(t_2 +t_1 ) [θ=60] t_2 +t_1 =((4(√(gh)))/g).((√3)/2)= t_2 +t_1 =2(√(((3h)/g) ))

$$\mathrm{x}−\mathrm{direction} \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{v}_{\mathrm{o}} \mathrm{cos}\:\theta\mathrm{t}_{\mathrm{1}} \\ $$$$\mathrm{x}_{\mathrm{2}} =\mathrm{v}_{\mathrm{o}} \mathrm{cos}\:\theta\mathrm{t}_{\mathrm{2}} \\ $$$$\mathrm{v}_{\mathrm{o}} =\mathrm{2}\sqrt{\mathrm{gh}} \\ $$$$\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}} =\mathrm{2h}\:\:\:\:\:\:\left(\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} \right)=\mathrm{2}\sqrt{\frac{\mathrm{h}}{\mathrm{g}}} \\ $$$$\mathrm{v}_{\mathrm{o}} \mathrm{cos}\:\theta\left(\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} \right)=\mathrm{2h} \\ $$$$\mathrm{cos}\theta=\frac{\mathrm{2h}}{\mathrm{v}_{\mathrm{o}} \left(\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} \right)}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:\: \\ $$$$\theta=\:\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\mathrm{60}^{°} \\ $$$$\mathrm{also}\:\:\mathrm{y}−\mathrm{direction} \\ $$$$\mathrm{y}=\mathrm{v}_{\mathrm{o}} \mathrm{sin}\:\theta\mathrm{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{y}=\mathrm{v}_{\mathrm{o}} \mathrm{sin}\:\theta\mathrm{t}_{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{v}_{\mathrm{o}} \mathrm{sin}\:\theta\mathrm{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{v}_{\mathrm{o}} \mathrm{sin}\:\theta\mathrm{t}_{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\mathrm{gh}}\:\mathrm{sin}\:\theta\left(\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$$\mathrm{2}\sqrt{\mathrm{gh}}\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}_{\mathrm{2}} +\mathrm{t}_{\mathrm{1}} \right)\:\:\:\left[\theta=\mathrm{60}\right] \\ $$$$\mathrm{t}_{\mathrm{2}} +\mathrm{t}_{\mathrm{1}} =\frac{\mathrm{4}\sqrt{\mathrm{gh}}}{\mathrm{g}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}= \\ $$$$\mathrm{t}_{\mathrm{2}} +\mathrm{t}_{\mathrm{1}} =\mathrm{2}\sqrt{\frac{\mathrm{3h}}{\mathrm{g}}\:}\: \\ $$$$ \\ $$$$ \\ $$

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