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Question Number 217575 by ajfour last updated on 16/Mar/25
Answered by mahdipoor last updated on 16/Mar/25
u_i , θ_i ⇒ d_i =((2u_i ^2 sin(θ_i )cos(θ_i ))/g) u_(i+1) ^→ = u_i (cos(θ_i )i+esin(θ_i )j) = u_(i+1) , θ_(i+1) { ((u_(i+1) ^2 =u_i ^2 (cos^2 (θ_i )+e^2 sin^2 (θ_i )))),((sin(θ_(i+1) )cos(θ_(i+1) )=((esin(θ_i )cos(θ_i ))/(cos^2 (θ_i )+e^2 sin^2 (θ_i ))))) :} d_(i+1) = ed_i Σd=(1+e+e^2 +...)d_0 =((1−e^(oo) )/(1−e))d_0 ⇒ e<1 ⇒ (1/(1−e))×((u_0 ^2 sin(2θ_0 ))/g)
$${u}_{{i}} \:,\:\theta_{{i}} \:\Rightarrow \\ $$$${d}_{{i}} =\frac{\mathrm{2}{u}_{{i}} ^{\mathrm{2}} {sin}\left(\theta_{{i}} \right){cos}\left(\theta_{{i}} \right)}{{g}} \\ $$$$\overset{\rightarrow} {{u}}_{{i}+\mathrm{1}} =\:{u}_{{i}} \left({cos}\left(\theta_{{i}} \right)\boldsymbol{\mathrm{i}}+{esin}\left(\theta_{{i}} \right)\boldsymbol{\mathrm{j}}\right)\:=\:{u}_{{i}+\mathrm{1}} \:,\:\theta_{{i}+\mathrm{1}} \\ $$$$\begin{cases}{{u}_{{i}+\mathrm{1}} ^{\mathrm{2}} ={u}_{{i}} ^{\mathrm{2}} \left({cos}^{\mathrm{2}} \left(\theta_{{i}} \right)+{e}^{\mathrm{2}} {sin}^{\mathrm{2}} \left(\theta_{{i}} \right)\right)}\\{{sin}\left(\theta_{{i}+\mathrm{1}} \right){cos}\left(\theta_{{i}+\mathrm{1}} \right)=\frac{{esin}\left(\theta_{{i}} \right){cos}\left(\theta_{{i}} \right)}{{cos}^{\mathrm{2}} \left(\theta_{{i}} \right)+{e}^{\mathrm{2}} {sin}^{\mathrm{2}} \left(\theta_{{i}} \right)}}\end{cases} \\ $$$${d}_{{i}+\mathrm{1}} =\:{ed}_{{i}} \\ $$$$\Sigma{d}=\left(\mathrm{1}+{e}+{e}^{\mathrm{2}} +…\right){d}_{\mathrm{0}} =\frac{\mathrm{1}−{e}^{{oo}} }{\mathrm{1}−{e}}{d}_{\mathrm{0}} \:\Rightarrow\:{e}<\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{e}}×\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} {sin}\left(\mathrm{2}\theta_{\mathrm{0}} \right)}{{g}} \\ $$
Commented by ajfour last updated on 16/Mar/25
example u=8m/s g=10m/s^2 θ_0 =45° e=(1/2) d_0 =((64)/(10))=6.4m Σd=((2×64)/(10))=12.8 m Thank you. i ll review again.
$${example} \\ $$$${u}=\mathrm{8}{m}/{s} \\ $$$${g}=\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$$$\theta_{\mathrm{0}} =\mathrm{45}° \\ $$$${e}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${d}_{\mathrm{0}} =\frac{\mathrm{64}}{\mathrm{10}}=\mathrm{6}.\mathrm{4}{m} \\ $$$$\Sigma{d}=\frac{\mathrm{2}×\mathrm{64}}{\mathrm{10}}=\mathrm{12}.\mathrm{8}\:{m} \\ $$$${Thank}\:{you}.\:{i}\:{ll}\:{review}\:{again}. \\ $$
Answered by mr W last updated on 16/Mar/25
Commented by mr W last updated on 17/Mar/25
0<e<1 u_(n−1) cos θ_(n−1) =u_n cos θ_n ...(i) eu_(n−1) sin θ_(n−1) =u_n sin θ_n ...(ii) (u_n /u_(n−1) )=((cos θ_(n−1) )/(cos θ_n )) ((tan θ_n )/(tan θ_(n−1) ))=e l_n =((2u_n ^2 sin θ_n cos θ_n )/g) l_(n−1) =((2u_(n−1) ^2 sin θ_(n−1) cos θ_(n−1) )/g) (l_n /l_(n−1) )=((u_n /u_(n−1) ))^2 ((sin θ_n cos θ_n )/(sin θ_(n−1) cos θ_(n−1) )) =((cos^2 θ_(n−1) sin θ_n cos θ_n )/(cos^2 θ_n sin θ_(n−1) cos θ_(n−1) )) =((tan θ_n )/(tan θ_(n−1) ))=e l_n =el_(n−1) =e^2 l_(n−2) =...=e^n l_0 with l_0 =((u_0 ^2 sin 2θ_0 )/g) L=Σ_(n=0) ^∞ l_n =l_0 Σ_(n=0) ^∞ e^n =(l_0 /(1−e))=((u_0 ^2 sin 2θ_0 )/(g(1−e)))
$$\mathrm{0}<{e}<\mathrm{1} \\ $$$${u}_{{n}−\mathrm{1}} \mathrm{cos}\:\theta_{{n}−\mathrm{1}} ={u}_{{n}} \:\mathrm{cos}\:\theta_{{n}} \:\:\:…\left({i}\right) \\ $$$${eu}_{{n}−\mathrm{1}} \mathrm{sin}\:\theta_{{n}−\mathrm{1}} ={u}_{{n}} \:\mathrm{sin}\:\theta_{{n}} \:\:\:…\left({ii}\right) \\ $$$$\frac{{u}_{{n}} }{{u}_{{n}−\mathrm{1}} }=\frac{\mathrm{cos}\:\theta_{{n}−\mathrm{1}} }{\mathrm{cos}\:\theta_{{n}} } \\ $$$$\frac{\mathrm{tan}\:\theta_{{n}} }{\mathrm{tan}\:\theta_{{n}−\mathrm{1}} }={e} \\ $$$$ \\ $$$${l}_{{n}} =\frac{\mathrm{2}{u}_{{n}} ^{\mathrm{2}} \:\mathrm{sin}\:\theta_{{n}} \:\mathrm{cos}\:\theta_{{n}} }{{g}} \\ $$$${l}_{{n}−\mathrm{1}} =\frac{\mathrm{2}{u}_{{n}−\mathrm{1}} ^{\mathrm{2}} \:\mathrm{sin}\:\theta_{{n}−\mathrm{1}} \:\mathrm{cos}\:\theta_{{n}−\mathrm{1}} }{{g}} \\ $$$$\frac{{l}_{{n}} }{{l}_{{n}−\mathrm{1}} }=\left(\frac{{u}_{{n}} }{{u}_{{n}−\mathrm{1}} }\right)^{\mathrm{2}} \frac{\mathrm{sin}\:\theta_{{n}} \:\mathrm{cos}\:\theta_{{n}} }{\mathrm{sin}\:\theta_{{n}−\mathrm{1}} \:\mathrm{cos}\:\theta_{{n}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta_{{n}−\mathrm{1}} \:\mathrm{sin}\:\theta_{{n}} \:\mathrm{cos}\:\theta_{{n}} }{\mathrm{cos}^{\mathrm{2}} \:\theta_{{n}} \:\mathrm{sin}\:\theta_{{n}−\mathrm{1}} \:\mathrm{cos}\:\theta_{{n}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{tan}\:\theta_{{n}} }{\mathrm{tan}\:\theta_{{n}−\mathrm{1}} }={e} \\ $$$${l}_{{n}} ={el}_{{n}−\mathrm{1}} ={e}^{\mathrm{2}} {l}_{{n}−\mathrm{2}} =…={e}^{{n}} {l}_{\mathrm{0}} \\ $$$${with}\:{l}_{\mathrm{0}} =\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta_{\mathrm{0}} }{{g}} \\ $$$${L}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{l}_{{n}} ={l}_{\mathrm{0}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{{n}} =\frac{{l}_{\mathrm{0}} }{\mathrm{1}−{e}}=\frac{{u}_{\mathrm{0}} ^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta_{\mathrm{0}} }{{g}\left(\mathrm{1}−{e}\right)} \\ $$

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