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Question-197619

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Question Number 197619 by mr W last updated on 24/Sep/23
Commented by mr W last updated on 24/Sep/23
assume the hill has the shape of a parabola. find the minimum speed with which a projectile should be launched from point C such that it can hit point B.
$${assume}\:{the}\:{hill}\:{has}\:{the}\:{shape}\:{of}\:{a}\: \\ $$$${parabola}.\:{find}\:{the}\:{minimum}\:{speed} \\ $$$${with}\:{which}\:{a}\:{projectile}\:{should}\:{be} \\ $$$${launched}\:{from}\:{point}\:{C}\:{such}\:{that}\:{it} \\ $$$${can}\:{hit}\:{point}\:{B}. \\ $$
Answered by mahdipoor last updated on 24/Sep/23
position of ball ⇒ { ((x=ut.cosβ)),((y=ut.sinβ−((gt^2 )/2)=x.tanβ−((gx^2 )/(2u^2 cos^2 β)))) :} I) when x=b+2a⇒y=0 0=(b+2a)[tanβ−((g(b+2a))/(2u^2 cos^2 β))]⇒(g/(2u^2 cos^2 β))=((tanβ)/(b+2a)) ⇒u=(√(((g(b+2a))/2)((1/(tanβ))+tanβ))) II) 1}h−y=(h/a^2 )(x−(a+b))^2 2}y=x.tanβ−((gx^2 )/(2u^2 cos^2 β))=x.tanβ[1−(x/(b+2a))] 1,2 ⇒(((tanβ)/(b+2a))−(h/a^2 ))x^2 +(((2h(a+b))/a^2 )−tanβ)x −((h(2ab+b^2 ))/a^2 )=0 ⇒((x.tanβ)/(b+2a))(x−(b+2a))−(h/a^2 )(x−(b+2a))(x−b)=0 ⇒(x−(b+2a))(x(((tanβ)/(b+2a))−(h/a^2 ))+((hb)/a^2 ))=0 ⇒^1 x_2 =((hb/a^2 )/(h/a^2 −tanβ/(b+2a)))≤0 or ∄⇒(h/a^2 )≤((tanβ)/(b+2a)) ⇒((h(b+2a))/a^2 )≤tanβ ⇒^2 x_2 =((hb/a^2 )/(h/a^2 −tanβ/(b+2a)))≥(b+2a)⇒((tanβ−2h/a)/(h/a^2 −tanβ/(b+2a)))≥0 ⇒ { ((((2h)/a)≤tanβ<((h(b+2a))/a^2 ))),((((2h)/a)≥tanβ>((h(b+2a))/a^2 ) impossible (2>2+(b/a)))) :} ⇒((2h)/a)≤tanβ<((2h(b+2a))/a^2 ) ⇒^(1,2) ((2h)/a)≤tanβ f=min((1/(tanβ))+tanβ) = { ((2 if ((2h)/a)≤1)),((((2h)/a)+(a/(2h)) if ((2h)/a)>1)) :} ⇒⇒⇒⇒min u=(√((g(b+2a)f)/2))
$${position}\:{of}\:{ball}\:\Rightarrow \\ $$$$\begin{cases}{{x}={ut}.{cos}\beta}\\{{y}={ut}.{sin}\beta−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}={x}.{tan}\beta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} {cos}^{\mathrm{2}} \beta}}\end{cases} \\ $$$$\left.\mathrm{I}\right)\:{when}\:\:\:{x}={b}+\mathrm{2}{a}\Rightarrow{y}=\mathrm{0} \\ $$$$\mathrm{0}=\left({b}+\mathrm{2}{a}\right)\left[{tan}\beta−\frac{{g}\left({b}+\mathrm{2}{a}\right)}{\mathrm{2}{u}^{\mathrm{2}} {cos}^{\mathrm{2}} \beta}\right]\Rightarrow\frac{{g}}{\mathrm{2}{u}^{\mathrm{2}} {cos}^{\mathrm{2}} \beta}=\frac{{tan}\beta}{{b}+\mathrm{2}{a}} \\ $$$$\Rightarrow{u}=\sqrt{\frac{{g}\left({b}+\mathrm{2}{a}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}}{{tan}\beta}+{tan}\beta\right)} \\ $$$$\left.\mathrm{II}\right)\:\: \\ $$$$\left.\mathrm{1}\right\}{h}−{y}=\frac{{h}}{{a}^{\mathrm{2}} }\left({x}−\left({a}+{b}\right)\right)^{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right\}{y}={x}.{tan}\beta−\frac{{gx}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} {cos}^{\mathrm{2}} \beta}={x}.{tan}\beta\left[\mathrm{1}−\frac{{x}}{{b}+\mathrm{2}{a}}\right] \\ $$$$\mathrm{1},\mathrm{2}\:\Rightarrow\left(\frac{{tan}\beta}{{b}+\mathrm{2}{a}}−\frac{{h}}{{a}^{\mathrm{2}} }\right){x}^{\mathrm{2}} +\left(\frac{\mathrm{2}{h}\left({a}+{b}\right)}{{a}^{\mathrm{2}} }−{tan}\beta\right){x} \\ $$$$−\frac{{h}\left(\mathrm{2}{ab}+{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{{x}.{tan}\beta}{{b}+\mathrm{2}{a}}\left({x}−\left({b}+\mathrm{2}{a}\right)\right)−\frac{{h}}{{a}^{\mathrm{2}} }\left({x}−\left({b}+\mathrm{2}{a}\right)\right)\left({x}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\left({b}+\mathrm{2}{a}\right)\right)\left({x}\left(\frac{{tan}\beta}{{b}+\mathrm{2}{a}}−\frac{{h}}{{a}^{\mathrm{2}} }\right)+\frac{{hb}}{{a}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\overset{\mathrm{1}} {\Rightarrow}{x}_{\mathrm{2}} =\frac{{hb}/{a}^{\mathrm{2}} }{{h}/{a}^{\mathrm{2}} −{tan}\beta/\left({b}+\mathrm{2}{a}\right)}\leqslant\mathrm{0}\:{or}\:\nexists\Rightarrow\frac{{h}}{{a}^{\mathrm{2}} }\leqslant\frac{{tan}\beta}{{b}+\mathrm{2}{a}} \\ $$$$\Rightarrow\frac{{h}\left({b}+\mathrm{2}{a}\right)}{{a}^{\mathrm{2}} }\leqslant{tan}\beta \\ $$$$\overset{\mathrm{2}} {\Rightarrow}{x}_{\mathrm{2}} =\frac{{hb}/{a}^{\mathrm{2}} }{{h}/{a}^{\mathrm{2}} −{tan}\beta/\left({b}+\mathrm{2}{a}\right)}\geqslant\left({b}+\mathrm{2}{a}\right)\Rightarrow\frac{{tan}\beta−\mathrm{2}{h}/{a}}{{h}/{a}^{\mathrm{2}} −{tan}\beta/\left({b}+\mathrm{2}{a}\right)}\geqslant\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\frac{\mathrm{2}{h}}{{a}}\leqslant{tan}\beta<\frac{{h}\left({b}+\mathrm{2}{a}\right)}{{a}^{\mathrm{2}} }}\\{\frac{\mathrm{2}{h}}{{a}}\geqslant{tan}\beta>\frac{{h}\left({b}+\mathrm{2}{a}\right)}{{a}^{\mathrm{2}} }\:\:\:{impossible}\:\left(\mathrm{2}>\mathrm{2}+\frac{{b}}{{a}}\right)}\end{cases}\:\: \\ $$$$\Rightarrow\frac{\mathrm{2}{h}}{{a}}\leqslant{tan}\beta<\frac{\mathrm{2}{h}\left({b}+\mathrm{2}{a}\right)}{{a}^{\mathrm{2}} } \\ $$$$\overset{\mathrm{1},\mathrm{2}} {\Rightarrow}\:\:\:\frac{\mathrm{2}{h}}{{a}}\leqslant{tan}\beta \\ $$$${f}={min}\left(\frac{\mathrm{1}}{{tan}\beta}+{tan}\beta\right)\:=\begin{cases}{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:\:\frac{\mathrm{2}{h}}{{a}}\leqslant\mathrm{1}}\\{\frac{\mathrm{2}{h}}{{a}}+\frac{{a}}{\mathrm{2}{h}}\:\:\:\:\:\:{if}\:\:\:\frac{\mathrm{2}{h}}{{a}}>\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\Rightarrow\Rightarrow\Rightarrow{min}\:{u}=\sqrt{\frac{{g}\left({b}+\mathrm{2}{a}\right){f}}{\mathrm{2}}} \\ $$
Commented by mr W last updated on 24/Sep/23
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 24/Sep/23
Commented by mr W last updated on 25/Sep/23
x=u cos θ t=2a+b ⇒t=((2a+b)/(u cos θ)) y=u sin θ t−((gt^2 )/2)=0 u sin θ−(g/2)×((2a+b)/(u cos θ))=0 ⇒u^2 =(((2a+b)g)/(sin 2θ)) φ=tan^(−1) ((2h)/a) θ≥φ=tan^(−1) ((2h)/a) if φ≤(π/4)=45°, i.e. (h/a)≤(1/2): u_(min) ^2 =(2a+b)g ⇒u_(min) =(√((2a+b)g)) if φ>(π/4)=45°, i.e. (h/a)>(1/2): u_(min) ^2 =(((2a+b)g)/(sin 2φ))=((a/(4h))+(h/a))(2a+b)g ⇒u_(min) =(√(((a/(4h))+(h/a))(2a+b)g))
$${x}={u}\:\mathrm{cos}\:\theta\:{t}=\mathrm{2}{a}+{b} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{a}+{b}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${y}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$${u}\:\mathrm{sin}\:\theta−\frac{{g}}{\mathrm{2}}×\frac{\mathrm{2}{a}+{b}}{{u}\:\mathrm{cos}\:\theta}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\frac{\left(\mathrm{2}{a}+{b}\right){g}}{\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\phi=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{h}}{{a}} \\ $$$$\theta\geqslant\phi=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{h}}{{a}} \\ $$$${if}\:\phi\leqslant\frac{\pi}{\mathrm{4}}=\mathrm{45}°,\:{i}.{e}.\:\frac{{h}}{{a}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${u}_{{min}} ^{\mathrm{2}} =\left(\mathrm{2}{a}+{b}\right){g} \\ $$$$\Rightarrow{u}_{{min}} =\sqrt{\left(\mathrm{2}{a}+{b}\right){g}} \\ $$$${if}\:\phi>\frac{\pi}{\mathrm{4}}=\mathrm{45}°,\:{i}.{e}.\:\frac{{h}}{{a}}>\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${u}_{{min}} ^{\mathrm{2}} =\frac{\left(\mathrm{2}{a}+{b}\right){g}}{\mathrm{sin}\:\mathrm{2}\phi}=\left(\frac{{a}}{\mathrm{4}{h}}+\frac{{h}}{{a}}\right)\left(\mathrm{2}{a}+{b}\right){g} \\ $$$$\Rightarrow{u}_{{min}} =\sqrt{\left(\frac{{a}}{\mathrm{4}{h}}+\frac{{h}}{{a}}\right)\left(\mathrm{2}{a}+{b}\right){g}} \\ $$
Commented by mahdipoor last updated on 25/Sep/23
(1/(sin2φ))=(1/(2tanφ.cos^2 ∅))=((1+tan^2 φ)/(2tan∅))=(1/2)((1/(tan∅))+tanφ) ⇒tanφ=((2h)/a)⇒(1/(sin2φ))=(h/a)+(a/(4h)) !?
$$\frac{\mathrm{1}}{{sin}\mathrm{2}\phi}=\frac{\mathrm{1}}{\mathrm{2}{tan}\phi.{cos}^{\mathrm{2}} \emptyset}=\frac{\mathrm{1}+{tan}^{\mathrm{2}} \phi}{\mathrm{2}{tan}\emptyset}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{tan}\emptyset}+{tan}\phi\right) \\ $$$$\Rightarrow{tan}\phi=\frac{\mathrm{2}{h}}{{a}}\Rightarrow\frac{\mathrm{1}}{{sin}\mathrm{2}\phi}=\frac{{h}}{{a}}+\frac{{a}}{\mathrm{4}{h}}\:\:!? \\ $$
Commented by mr W last updated on 25/Sep/23
yes, you are right sir!
$${yes},\:{you}\:{are}\:{right}\:{sir}! \\ $$

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