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Question Number 59301 by ajfour last updated on 07/May/19
Commented by ajfour last updated on 07/May/19
If deceleration of bullet through wall be α, find y(<h) for maximum x. ...
$$\mathrm{If}\:\mathrm{deceleration}\:\mathrm{of}\:\mathrm{bullet}\:\mathrm{through} \\ $$$$\:\mathrm{wall}\:\mathrm{be}\:\alpha,\:\mathrm{find}\:\mathrm{y}\left(<\mathrm{h}\right)\:\mathrm{for}\:\mathrm{maximum}\:\mathrm{x}. \\ $$$$… \\ $$
Commented by tanmay last updated on 07/May/19
Commented by tanmay last updated on 07/May/19
v^2 =u^2 −2α(a+2b) tan30^o =(b/y) b=(y/( (√3))) v^2 =u^2 −2α(a+((2y)/( (√3)))) y=(1/2)gt^2 x=vt x=v×(√((2y)/g)) x=[u^2 −2α(a+((2y)/( (√3))))]^(1/2) ×(√((2y)/g)) x^2 =[u^2 −2α(a+((2y)/( (√3))))]×((2y)/g) x^2 =(u^2 −2α)×((2y)/g)−((4αy)/( (√3)))×((2y)/g) x^2 =Ay−By^2 (x^2 /B)=(A/B)y−y^2 (x^2 /B)=−(y^2 −2×y×(A/(2B))+(A^2 /(4B^2 ))−(A^2 /(4B^2 ))) (x^2 /B)=(A^2 /(4B^2 ))−(y−(A/(2B)))^2 x_(max) =(A^2 /(4B))=(([(2/g)(u^2 −2α)]^2 )/(4×((8α)/( (√3) g)) )) x_(max) =(4/g^2 )×(((u^2 −2α)^2 )/(32α))×(√3) g x_(max) =(((√3) (u^2 −2α)^2 )/(8g×α)) when y=(A/(2B))=(((2/g)(u^2 −2α))/((8α)/( (√3) g)))=((2(u^2 −2α)×(√3) )/(8α))
$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}\alpha\left({a}+\mathrm{2}{b}\right) \\ $$$${tan}\mathrm{30}^{{o}} =\frac{{b}}{{y}} \\ $$$${b}=\frac{{y}}{\:\sqrt{\mathrm{3}}} \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}\alpha\left({a}+\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${x}={vt} \\ $$$${x}={v}×\sqrt{\frac{\mathrm{2}{y}}{{g}}}\: \\ $$$${x}=\left[{u}^{\mathrm{2}} −\mathrm{2}\alpha\left({a}+\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{3}}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} ×\sqrt{\frac{\mathrm{2}{y}}{{g}}}\: \\ $$$${x}^{\mathrm{2}} =\left[{u}^{\mathrm{2}} −\mathrm{2}\alpha\left({a}+\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{3}}}\right)\right]×\frac{\mathrm{2}{y}}{{g}} \\ $$$${x}^{\mathrm{2}} =\left({u}^{\mathrm{2}} −\mathrm{2}\alpha\right)×\frac{\mathrm{2}{y}}{{g}}−\frac{\mathrm{4}\alpha{y}}{\:\sqrt{\mathrm{3}}}×\frac{\mathrm{2}{y}}{{g}} \\ $$$${x}^{\mathrm{2}} ={Ay}−{By}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{{B}}=\frac{{A}}{{B}}{y}−{y}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{{B}}=−\left({y}^{\mathrm{2}} −\mathrm{2}×{y}×\frac{{A}}{\mathrm{2}{B}}+\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}^{\mathrm{2}} }−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}^{\mathrm{2}} }\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{{B}}=\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}^{\mathrm{2}} }−\left({y}−\frac{{A}}{\mathrm{2}{B}}\right)^{\mathrm{2}} \\ $$$${x}_{{max}} =\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}=\frac{\left[\frac{\mathrm{2}}{{g}}\left({u}^{\mathrm{2}} −\mathrm{2}\alpha\right)\right]^{\mathrm{2}} }{\mathrm{4}×\frac{\mathrm{8}\alpha}{\:\sqrt{\mathrm{3}}\:{g}}\:} \\ $$$${x}_{{max}} =\frac{\mathrm{4}}{{g}^{\mathrm{2}} }×\frac{\left({u}^{\mathrm{2}} −\mathrm{2}\alpha\right)^{\mathrm{2}} }{\mathrm{32}\alpha}×\sqrt{\mathrm{3}}\:{g} \\ $$$${x}_{{max}} =\frac{\sqrt{\mathrm{3}}\:\left({u}^{\mathrm{2}} −\mathrm{2}\alpha\right)^{\mathrm{2}} }{\mathrm{8}{g}×\alpha} \\ $$$${when}\:{y}=\frac{{A}}{\mathrm{2}{B}}=\frac{\frac{\mathrm{2}}{{g}}\left({u}^{\mathrm{2}} −\mathrm{2}\alpha\right)}{\frac{\mathrm{8}\alpha}{\:\sqrt{\mathrm{3}}\:{g}}}=\frac{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{2}\alpha\right)×\sqrt{\mathrm{3}}\:}{\mathrm{8}\alpha} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by ajfour last updated on 08/May/19
let me sometime, thanks Sir.
$$\mathrm{let}\:\mathrm{me}\:\mathrm{sometime},\:\mathrm{thanks}\:\mathrm{Sir}. \\ $$
Answered by mr W last updated on 08/May/19
s=path length in wall v=outgoing velocity s=ut_1 −(1/2)αt_1 ^2 v=u−αt_1 v^2 =u^2 −2αut_1 +α^2 ut_1 ^2 =u^2 −2α(ut_1 −(1/2)αt_1 ^2 ) v^2 =u^2 −2αs v=(√(u^2 −2αs)) s=a+((2y)/( (√3))) ⇒v=(√(u^2 −2α(a+((2y)/( (√3))))))=(√(u^2 −2αa−((4αy)/( (√3))))) y=(1/2)gt_2 ^2 x=(y/( (√3)))+vt_2 =(y/( (√3)))+v(√((2y)/g)) ⇒x=(y/( (√3)))+(√(((2y)/g)(u^2 −2αa−((4αy)/( (√3)))))) let λ=(y/( (√3))) ⇒x=λ+(√(((8α(√3))/g)(((u^2 −2αa)/(4α))−λ)λ)) ⇒x=λ+(√(δ(μ−λ)λ)) with δ=((8α(√3))/g), μ=((u^2 −2αa)/(4α))=(u^2 /(4α))−(a/2) (dx/dλ)=1+((δ(μ−2λ))/(2(√(δ(μ−λ)λ))))=0 ((δ(2λ−μ))/(2(√(δ(μ−λ)λ))))=1 δ^2 (μ^2 −4μλ+4λ^2 )=4δ(μ−λ)λ ⇒4(δ−1)λ^2 −4μ(δ+1)λ+δμ^2 =0 ⇒λ=((μ(δ+1−(√(3δ+1))))/(2(δ−1))) ⇒y=(√3)λ=((μ(√3)(δ+1−(√(3δ+1))))/(2(δ−1)))
$${s}={path}\:{length}\:{in}\:{wall} \\ $$$${v}={outgoing}\:{velocity} \\ $$$${s}={ut}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\alpha{t}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${v}={u}−\alpha{t}_{\mathrm{1}} \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}\alpha{ut}_{\mathrm{1}} +\alpha^{\mathrm{2}} {ut}_{\mathrm{1}} ^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}\alpha\left({ut}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\alpha{t}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}\alpha{s} \\ $$$${v}=\sqrt{{u}^{\mathrm{2}} −\mathrm{2}\alpha{s}} \\ $$$${s}={a}+\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{v}=\sqrt{{u}^{\mathrm{2}} −\mathrm{2}\alpha\left({a}+\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{3}}}\right)}=\sqrt{{u}^{\mathrm{2}} −\mathrm{2}\alpha{a}−\frac{\mathrm{4}\alpha{y}}{\:\sqrt{\mathrm{3}}}} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{gt}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${x}=\frac{{y}}{\:\sqrt{\mathrm{3}}}+{vt}_{\mathrm{2}} =\frac{{y}}{\:\sqrt{\mathrm{3}}}+{v}\sqrt{\frac{\mathrm{2}{y}}{{g}}} \\ $$$$\Rightarrow{x}=\frac{{y}}{\:\sqrt{\mathrm{3}}}+\sqrt{\frac{\mathrm{2}{y}}{{g}}\left({u}^{\mathrm{2}} −\mathrm{2}\alpha{a}−\frac{\mathrm{4}\alpha{y}}{\:\sqrt{\mathrm{3}}}\right)} \\ $$$${let}\:\lambda=\frac{{y}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\lambda+\sqrt{\frac{\mathrm{8}\alpha\sqrt{\mathrm{3}}}{{g}}\left(\frac{{u}^{\mathrm{2}} −\mathrm{2}\alpha{a}}{\mathrm{4}\alpha}−\lambda\right)\lambda} \\ $$$$\Rightarrow{x}=\lambda+\sqrt{\delta\left(\mu−\lambda\right)\lambda} \\ $$$${with}\:\delta=\frac{\mathrm{8}\alpha\sqrt{\mathrm{3}}}{{g}},\:\mu=\frac{{u}^{\mathrm{2}} −\mathrm{2}\alpha{a}}{\mathrm{4}\alpha}=\frac{{u}^{\mathrm{2}} }{\mathrm{4}\alpha}−\frac{{a}}{\mathrm{2}} \\ $$$$\frac{{dx}}{{d}\lambda}=\mathrm{1}+\frac{\delta\left(\mu−\mathrm{2}\lambda\right)}{\mathrm{2}\sqrt{\delta\left(\mu−\lambda\right)\lambda}}=\mathrm{0} \\ $$$$\frac{\delta\left(\mathrm{2}\lambda−\mu\right)}{\mathrm{2}\sqrt{\delta\left(\mu−\lambda\right)\lambda}}=\mathrm{1} \\ $$$$\delta^{\mathrm{2}} \left(\mu^{\mathrm{2}} −\mathrm{4}\mu\lambda+\mathrm{4}\lambda^{\mathrm{2}} \right)=\mathrm{4}\delta\left(\mu−\lambda\right)\lambda \\ $$$$\Rightarrow\mathrm{4}\left(\delta−\mathrm{1}\right)\lambda^{\mathrm{2}} −\mathrm{4}\mu\left(\delta+\mathrm{1}\right)\lambda+\delta\mu^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mu\left(\delta+\mathrm{1}−\sqrt{\mathrm{3}\delta+\mathrm{1}}\right)}{\mathrm{2}\left(\delta−\mathrm{1}\right)} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{3}}\lambda=\frac{\mu\sqrt{\mathrm{3}}\left(\delta+\mathrm{1}−\sqrt{\mathrm{3}\delta+\mathrm{1}}\right)}{\mathrm{2}\left(\delta−\mathrm{1}\right)} \\ $$
Commented by ajfour last updated on 10/May/19
Thanks Sir, you have managed it too well. I shall also try..
$$\mathrm{Thanks}\:\mathrm{Sir},\:\mathrm{you}\:\mathrm{have}\:\mathrm{managed}\:\mathrm{it} \\ $$$$\mathrm{too}\:\mathrm{well}.\:\:\mathrm{I}\:\mathrm{shall}\:\mathrm{also}\:\mathrm{try}.. \\ $$

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