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Question Number 82057 by ajfour last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
The plane diving at speed u drops a ball which upon one bounce collides with the plane when it reaches its maximum height. Find the declination of u^� from horizontal.
$${The}\:{plane}\:{diving}\:{at}\:{speed}\:{u}\:{drops} \\ $$$${a}\:{ball}\:{which}\:{upon}\:{one}\:{bounce} \\ $$$${collides}\:{with}\:{the}\:{plane}\:{when} \\ $$$${it}\:{reaches}\:{its}\:{maximum}\:{height}. \\ $$$${Find}\:{the}\:{declination}\:{of}\:\bar {{u}}\:{from} \\ $$$${horizontal}. \\ $$$$ \\ $$
Answered by mr W last updated on 18/Feb/20
Commented by mr W last updated on 18/Feb/20
e(√(2gh+(u sin θ)^2 ))=(√(2gh_2 )) ⇒h_2 =e^2 (1+((u^2 sin^2 θ)/(2gh)))h t=(((√(u^2 sin^2 θ+2gh))−u sin θ+(√(2gh_2 )))/g) t=(L/(u cos θ))=(h_1 /(tan θ u cos θ))=((h−h_2 )/(u sin θ)) ⇒L=(1/(tan θ))(1−e^2 −((e^2 u^2 sin^2 θ)/(2gh)))h (((√(u^2 sin^2 θ+2gh))−u sin θ+(√(2gh_2 )))/g)=(h/(u sin θ))−(h_2 /(u sin θ)) (√(u^2 sin^2 θ+2gh))−u sin θ+(√(2ghe^2 (1+((u^2 sin^2 θ)/(2gh)))))=((gh)/(u sin θ))−((gh)/(u sin θ))e^2 (1+((u^2 sin^2 θ)/(2gh))) (1+e)(√(u^2 sin^2 θ+2gh))=(((1−e^2 )gh)/(u sin θ))+(((2−e^2 )u sin θ)/2) 2(1+e)(√(1+((2gh)/(u^2 sin^2 θ))))=(((1−e^2 )2gh)/(u^2 sin^2 θ))+2−e^2 let λ=((2gh)/(u^2 sin^2 θ)) 2(1+e)(√(1+λ))=(1−e^2 )λ+2−e^2 (1−e^2 )^2 λ^2 −2e(1+e)(4+e−e^2 )λ−e(8+7e−e^3 )=0 λ=((e(4+e−e^2 )+(√(e(8+7e+2e^2 −e^3 ))))/((1+e)(1−e)^2 )) ⇒((2gh)/(u^2 sin^2 θ))=((e(4+e−e^2 )+(√(e(8+7e+2e^2 −e^3 ))))/((1+e)(1−e)^2 )) ⇒θ=sin^(−1) [(((1−e)(√(1+e)))/( (√(e(4+e−e^2 )+(√(e(8+7e+2e^2 −e^3 )))))))(√((2gh)/u^2 ))] example: e=0.5 θ≈sin^(−1) (0.286716(√((2gh)/u^2 )))
$${e}\sqrt{\mathrm{2}{gh}+\left({u}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}{gh}_{\mathrm{2}} } \\ $$$$\Rightarrow{h}_{\mathrm{2}} ={e}^{\mathrm{2}} \left(\mathrm{1}+\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{gh}}\right){h} \\ $$$${t}=\frac{\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gh}}−{u}\:\mathrm{sin}\:\theta+\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }}{{g}} \\ $$$${t}=\frac{{L}}{{u}\:\mathrm{cos}\:\theta}=\frac{{h}_{\mathrm{1}} }{\mathrm{tan}\:\theta\:{u}\:\mathrm{cos}\:\theta}=\frac{{h}−{h}_{\mathrm{2}} }{{u}\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{L}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\left(\mathrm{1}−{e}^{\mathrm{2}} −\frac{{e}^{\mathrm{2}} {u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{gh}}\right){h} \\ $$$$ \\ $$$$\frac{\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gh}}−{u}\:\mathrm{sin}\:\theta+\sqrt{\mathrm{2}{gh}_{\mathrm{2}} }}{{g}}=\frac{{h}}{{u}\:\mathrm{sin}\:\theta}−\frac{{h}_{\mathrm{2}} }{{u}\:\mathrm{sin}\:\theta} \\ $$$$\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gh}}−{u}\:\mathrm{sin}\:\theta+\sqrt{\mathrm{2}{ghe}^{\mathrm{2}} \left(\mathrm{1}+\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{gh}}\right)}=\frac{{gh}}{{u}\:\mathrm{sin}\:\theta}−\frac{{gh}}{{u}\:\mathrm{sin}\:\theta}{e}^{\mathrm{2}} \left(\mathrm{1}+\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}{gh}}\right) \\ $$$$\left(\mathrm{1}+{e}\right)\sqrt{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gh}}=\frac{\left(\mathrm{1}−{e}^{\mathrm{2}} \right){gh}}{{u}\:\mathrm{sin}\:\theta}+\frac{\left(\mathrm{2}−{e}^{\mathrm{2}} \right){u}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{1}+{e}\right)\sqrt{\mathrm{1}+\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}}=\frac{\left(\mathrm{1}−{e}^{\mathrm{2}} \right)\mathrm{2}{gh}}{{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}+\mathrm{2}−{e}^{\mathrm{2}} \\ $$$${let}\:\lambda=\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{2}\left(\mathrm{1}+{e}\right)\sqrt{\mathrm{1}+\lambda}=\left(\mathrm{1}−{e}^{\mathrm{2}} \right)\lambda+\mathrm{2}−{e}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{e}^{\mathrm{2}} \right)^{\mathrm{2}} \lambda^{\mathrm{2}} −\mathrm{2}{e}\left(\mathrm{1}+{e}\right)\left(\mathrm{4}+{e}−{e}^{\mathrm{2}} \right)\lambda−{e}\left(\mathrm{8}+\mathrm{7}{e}−{e}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\lambda=\frac{{e}\left(\mathrm{4}+{e}−{e}^{\mathrm{2}} \right)+\sqrt{{e}\left(\mathrm{8}+\mathrm{7}{e}+\mathrm{2}{e}^{\mathrm{2}} −{e}^{\mathrm{3}} \right)}}{\left(\mathrm{1}+{e}\right)\left(\mathrm{1}−{e}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}=\frac{{e}\left(\mathrm{4}+{e}−{e}^{\mathrm{2}} \right)+\sqrt{{e}\left(\mathrm{8}+\mathrm{7}{e}+\mathrm{2}{e}^{\mathrm{2}} −{e}^{\mathrm{3}} \right)}}{\left(\mathrm{1}+{e}\right)\left(\mathrm{1}−{e}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \left[\frac{\left(\mathrm{1}−{e}\right)\sqrt{\mathrm{1}+{e}}}{\:\sqrt{{e}\left(\mathrm{4}+{e}−{e}^{\mathrm{2}} \right)+\sqrt{{e}\left(\mathrm{8}+\mathrm{7}{e}+\mathrm{2}{e}^{\mathrm{2}} −{e}^{\mathrm{3}} \right)}}}\sqrt{\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} }}\right] \\ $$$$ \\ $$$${example}:\:{e}=\mathrm{0}.\mathrm{5} \\ $$$$\theta\approx\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{286716}\sqrt{\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} }}\right) \\ $$
Commented by ajfour last updated on 18/Feb/20
Very Nice, Sir. Awfully Good!
$$\mathcal{V}{ery}\:\mathcal{N}{ice},\:{Sir}.\:\mathcal{A}{wfully}\:{Good}! \\ $$
Commented by mr W last updated on 18/Feb/20
Commented by mr W last updated on 18/Feb/20
Commented by mr W last updated on 18/Feb/20
Answered by ajfour last updated on 18/Feb/20
tan θ=((h−H)/(a+b)) .....(i) e^2 v_y ^2 =e^2 (u^2 sin^2 θ+2gh) H=((e^2 v_y ^2 )/(2g))=((e^2 (u^2 sin^2 θ+2gh))/(2g)) (usin θ)t_1 +((gt_1 ^2 )/2)=h ⇒ t_1 =−((usin θ)/g)+(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) & t_2 =((ev_y )/g)=e(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) a+b=(ucos θ)(t_1 +t_2 ) from ..(i) (tan θ)(a+b)=h−H usin θ{−((usin θ)/g)+(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) +e(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) } = h−((e^2 (u^2 sin^2 θ+2gh))/(2g)) let ((usin θ)/g)=x, ((2h)/g)=c ⇒ x{−x+(√(x^2 +c))+e(√(x^2 +c)) } = (c/2)−(e^2 /2)(x^2 +c) ⇒ {x^2 (1−(e^2 /2))+(c/2)(1−e^2 )}^2 = x^2 (1+e)^2 (x^2 +c) For e=1/2 , also let x^2 /c = z {(7/8)z+(3/8)}^2 =(9/4)z(z+1) 49z^2 +42z+9=144z^2 +144z 95z^2 +102z−9=0 z=−((51)/(95))+(√((((51)/(95)))^2 +((9/(95))))) ≈0.08197636 ⇒ (((usin θ)/g))^2 =((2h)/g)×0.08197636 sin θ=((√(2gh))/u)×0.286315
$$\mathrm{tan}\:\theta=\frac{{h}−{H}}{{a}+{b}}\:\:\:\:\:…..\left({i}\right) \\ $$$${e}^{\mathrm{2}} {v}_{{y}} ^{\mathrm{2}} ={e}^{\mathrm{2}} \left({u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{2}{gh}\right) \\ $$$${H}=\frac{{e}^{\mathrm{2}} {v}_{{y}} ^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{{e}^{\mathrm{2}} \left({u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{2}{gh}\right)}{\mathrm{2}{g}} \\ $$$$\left({u}\mathrm{sin}\:\theta\right){t}_{\mathrm{1}} +\frac{{gt}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}={h} \\ $$$$\Rightarrow\:{t}_{\mathrm{1}} =−\frac{{u}\mathrm{sin}\:\theta}{{g}}+\sqrt{\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{g}^{\mathrm{2}} }+\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\:\&\:\:\:\:\:{t}_{\mathrm{2}} =\frac{{ev}_{{y}} }{{g}}={e}\sqrt{\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{g}^{\mathrm{2}} }+\frac{\mathrm{2}{h}}{{g}}} \\ $$$${a}+{b}=\left({u}\mathrm{cos}\:\theta\right)\left({t}_{\mathrm{1}} +{t}_{\mathrm{2}} \right) \\ $$$${from}\:..\left({i}\right) \\ $$$$\left(\mathrm{tan}\:\theta\right)\left({a}+{b}\right)={h}−{H} \\ $$$${u}\mathrm{sin}\:\theta\left\{−\frac{{u}\mathrm{sin}\:\theta}{{g}}+\sqrt{\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{g}^{\mathrm{2}} }+\frac{\mathrm{2}{h}}{{g}}}\right. \\ $$$$\left.\:\:\:\:\:\:+{e}\sqrt{\frac{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{g}^{\mathrm{2}} }+\frac{\mathrm{2}{h}}{{g}}}\:\right\} \\ $$$$\:\:\:\:\:=\:{h}−\frac{{e}^{\mathrm{2}} \left({u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{2}{gh}\right)}{\mathrm{2}{g}} \\ $$$${let}\:\:\:\frac{{u}\mathrm{sin}\:\theta}{{g}}={x},\:\:\frac{\mathrm{2}{h}}{{g}}={c}\:\:\Rightarrow \\ $$$$\:{x}\left\{−{x}+\sqrt{{x}^{\mathrm{2}} +{c}}+{e}\sqrt{{x}^{\mathrm{2}} +{c}}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{{c}}{\mathrm{2}}−\frac{{e}^{\mathrm{2}} }{\mathrm{2}}\left({x}^{\mathrm{2}} +{c}\right) \\ $$$$\Rightarrow \\ $$$$\left\{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{e}^{\mathrm{2}} }{\mathrm{2}}\right)+\frac{{c}}{\mathrm{2}}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\:{x}^{\mathrm{2}} \left(\mathrm{1}+{e}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{c}\right) \\ $$$${For}\:{e}=\mathrm{1}/\mathrm{2}\:,\:{also}\:{let}\:{x}^{\mathrm{2}} /{c}\:=\:{z} \\ $$$$\left\{\frac{\mathrm{7}}{\mathrm{8}}{z}+\frac{\mathrm{3}}{\mathrm{8}}\right\}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}{z}\left({z}+\mathrm{1}\right) \\ $$$$\mathrm{49}{z}^{\mathrm{2}} +\mathrm{42}{z}+\mathrm{9}=\mathrm{144}{z}^{\mathrm{2}} +\mathrm{144}{z} \\ $$$$\mathrm{95}{z}^{\mathrm{2}} +\mathrm{102}{z}−\mathrm{9}=\mathrm{0} \\ $$$${z}=−\frac{\mathrm{51}}{\mathrm{95}}+\sqrt{\left(\frac{\mathrm{51}}{\mathrm{95}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{9}}{\mathrm{95}}\right)} \\ $$$$\:\:\approx\mathrm{0}.\mathrm{08197636} \\ $$$$\:\:\Rightarrow\:\:\:\left(\frac{{u}\mathrm{sin}\:\theta}{{g}}\right)^{\mathrm{2}} =\frac{\mathrm{2}{h}}{{g}}×\mathrm{0}.\mathrm{08197636} \\ $$$$\:\:\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{2}{gh}}}{{u}}×\mathrm{0}.\mathrm{286315} \\ $$
Commented by ajfour last updated on 18/Feb/20
Yes sir, i agree, i shall reexamine this one later, i intend to try to obtain s_(max) , s_(min) of our eql.△ Q presently..
$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{agree},\:\mathrm{i}\:\mathrm{shall}\:\mathrm{reexamine} \\ $$$$\mathrm{this}\:\mathrm{one}\:\mathrm{later},\:\mathrm{i}\:\mathrm{intend}\:\mathrm{to}\:\mathrm{try}\:\mathrm{to} \\ $$$$\mathrm{obtain}\:\mathrm{s}_{\mathrm{max}} ,\:\mathrm{s}_{\mathrm{min}} \:\:\mathrm{of}\:\mathrm{our}\:\mathrm{eql}.\bigtriangleup\:\mathrm{Q} \\ $$$$\mathrm{presently}.. \\ $$
Commented by mr W last updated on 18/Feb/20
our results differ from each other. i can′t say where my or your error is. i graphed my result for various cases, both curves meet exactly on a point on the x−axis, so my result seems to be ok.
$${our}\:{results}\:{differ}\:{from}\:{each}\:{other}. \\ $$$${i}\:{can}'{t}\:{say}\:{where}\:{my}\:{or}\:{your}\:{error}\:{is}. \\ $$$${i}\:{graphed}\:{my}\:{result}\:{for}\:{various}\:{cases}, \\ $$$${both}\:{curves}\:{meet}\:{exactly}\:{on}\:{a}\:{point} \\ $$$${on}\:{the}\:{x}−{axis},\:{so}\:{my}\:{result}\:{seems}\:{to} \\ $$$${be}\:{ok}. \\ $$

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