Question Number 133838 by I want to learn more last updated on 24/Feb/21
Commented by mr W last updated on 25/Feb/21
$${it}\:{is}\:{not}\:{clear}\:{what}\:{is}\:{meant}.\:{is}\:{the} \\ $$$${lineman}\:\mathrm{4}.\mathrm{6}\:{away}\:{from}\:{the}\:{goalposts} \\ $$$${or}\:{from}\:{the}\:{kicker}?\:{i}\:{assumed}\:{he}\:{is} \\ $$$$\mathrm{4}.\mathrm{6}\:{m}\:{away}\:{from}\:{the}\:{goalposts}. \\ $$
Answered by mr W last updated on 25/Feb/21
Commented by I want to learn more last updated on 24/Feb/21
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$
Commented by mr W last updated on 24/Feb/21
![L=45.7 m h=3 m θ=45° (a) t=(L/(u cos θ)) h=(L/(u cos θ))(u sin θ−(g/2)×(L/(u cos θ))) h=L[tan θ−((gL)/(2u^2 ))(1+tan^2 θ)] (h/L)=tan θ−((gL)/(2u^2 ))(1+tan^2 θ) let λ=((gL)/(2u^2 )), η=(h/L), ξ=tan θ η=ξ−λ(1+ξ^2 ) λ=((ξ−η)/(1+ξ^2 )) (dλ/dξ)=(1/(1+ξ^2 ))−((2ξ^2 )/((1+ξ^2 )^2 ))=0 ⇒1−((2ξ^2 )/(1+ξ^2 ))=0 ⇒ξ^2 =1 ⇒tan θ=1 ⇒θ=45° i.e. at θ=45° we have λ_(max) or u_(min.) η=1−λ_(max) (1+1) ⇒λ_(max) =((1−η)/2) ((gL)/(2u_(min) ^2 ))=((1−(h/L))/2)=((L−h)/(2L)) u_(min) =(√((gL^2 )/(L−h)))=(√((10×45.7^2 )/(45.7−3)))=22.116 m/s (b) d=4.6 m h_1 =L_1 [tan θ−((gL_1 )/(2u^2 ))(1+tan^2 θ)] h_1 =L_1 [tan θ−(L_1 /L)×((gL)/(2u^2 ))(1+tan^2 θ)] h_1 =L_1 [1−(L_1 /L)×((L−h)/(2L))×2] h_1 =L_1 [1−(L_1 /L)×((L−h)/L)] at L_1 =L−d=45.7−4.6=41.1 m: h_1 =41.1×(1−((41.1×42.7)/(45.7^2 )))=6.56 m>2.5 m i.e. the lineman can′t block the ball. (c) d=1 m L_1 =44.7 m h_1 =44.7[1−((44.7×42.7)/(45.7^2 ))]=3.85 m > 2.5 m i.e. the lineman can′t block the ball.](Q133862.png)
$${L}=\mathrm{45}.\mathrm{7}\:{m} \\ $$$${h}=\mathrm{3}\:{m} \\ $$$$\theta=\mathrm{45}° \\ $$$$\left({a}\right) \\ $$$${t}=\frac{{L}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${h}=\frac{{L}}{{u}\:\mathrm{cos}\:\theta}\left({u}\:\mathrm{sin}\:\theta−\frac{{g}}{\mathrm{2}}×\frac{{L}}{{u}\:\mathrm{cos}\:\theta}\right) \\ $$$${h}={L}\left[\mathrm{tan}\:\theta−\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\right] \\ $$$$\frac{{h}}{{L}}=\mathrm{tan}\:\theta−\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right) \\ $$$${let}\:\lambda=\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} },\:\eta=\frac{{h}}{{L}},\:\xi=\mathrm{tan}\:\theta \\ $$$$\eta=\xi−\lambda\left(\mathrm{1}+\xi^{\mathrm{2}} \right) \\ $$$$\lambda=\frac{\xi−\eta}{\mathrm{1}+\xi^{\mathrm{2}} } \\ $$$$\frac{{d}\lambda}{{d}\xi}=\frac{\mathrm{1}}{\mathrm{1}+\xi^{\mathrm{2}} }−\frac{\mathrm{2}\xi^{\mathrm{2}} }{\left(\mathrm{1}+\xi^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}\xi^{\mathrm{2}} }{\mathrm{1}+\xi^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\xi^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\mathrm{tan}\:\theta=\mathrm{1}\:\Rightarrow\theta=\mathrm{45}° \\ $$$${i}.{e}.\:{at}\:\theta=\mathrm{45}°\:{we}\:{have}\:\lambda_{{max}} \:{or}\:{u}_{{min}.} \\ $$$$\eta=\mathrm{1}−\lambda_{{max}} \left(\mathrm{1}+\mathrm{1}\right)\:\Rightarrow\lambda_{{max}} =\frac{\mathrm{1}−\eta}{\mathrm{2}} \\ $$$$\frac{{gL}}{\mathrm{2}{u}_{{min}} ^{\mathrm{2}} }=\frac{\mathrm{1}−\frac{{h}}{{L}}}{\mathrm{2}}=\frac{{L}−{h}}{\mathrm{2}{L}} \\ $$$${u}_{{min}} =\sqrt{\frac{{gL}^{\mathrm{2}} }{{L}−{h}}}=\sqrt{\frac{\mathrm{10}×\mathrm{45}.\mathrm{7}^{\mathrm{2}} }{\mathrm{45}.\mathrm{7}−\mathrm{3}}}=\mathrm{22}.\mathrm{116}\:{m}/{s} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$${d}=\mathrm{4}.\mathrm{6}\:{m} \\ $$$${h}_{\mathrm{1}} ={L}_{\mathrm{1}} \left[\mathrm{tan}\:\theta−\frac{{gL}_{\mathrm{1}} }{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\right] \\ $$$${h}_{\mathrm{1}} ={L}_{\mathrm{1}} \left[\mathrm{tan}\:\theta−\frac{{L}_{\mathrm{1}} }{{L}}×\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\right] \\ $$$${h}_{\mathrm{1}} ={L}_{\mathrm{1}} \left[\mathrm{1}−\frac{{L}_{\mathrm{1}} }{{L}}×\frac{{L}−{h}}{\mathrm{2}{L}}×\mathrm{2}\right] \\ $$$${h}_{\mathrm{1}} ={L}_{\mathrm{1}} \left[\mathrm{1}−\frac{{L}_{\mathrm{1}} }{{L}}×\frac{{L}−{h}}{{L}}\right] \\ $$$${at}\:{L}_{\mathrm{1}} ={L}−{d}=\mathrm{45}.\mathrm{7}−\mathrm{4}.\mathrm{6}=\mathrm{41}.\mathrm{1}\:{m}: \\ $$$${h}_{\mathrm{1}} =\mathrm{41}.\mathrm{1}×\left(\mathrm{1}−\frac{\mathrm{41}.\mathrm{1}×\mathrm{42}.\mathrm{7}}{\mathrm{45}.\mathrm{7}^{\mathrm{2}} }\right)=\mathrm{6}.\mathrm{56}\:{m}>\mathrm{2}.\mathrm{5}\:{m} \\ $$$${i}.{e}.\:{the}\:{lineman}\:{can}'{t}\:{block}\:{the}\:{ball}. \\ $$$$ \\ $$$$\left({c}\right) \\ $$$${d}=\mathrm{1}\:{m} \\ $$$${L}_{\mathrm{1}} =\mathrm{44}.\mathrm{7}\:{m} \\ $$$${h}_{\mathrm{1}} =\mathrm{44}.\mathrm{7}\left[\mathrm{1}−\frac{\mathrm{44}.\mathrm{7}×\mathrm{42}.\mathrm{7}}{\mathrm{45}.\mathrm{7}^{\mathrm{2}} }\right]=\mathrm{3}.\mathrm{85}\:{m}\:>\:\mathrm{2}.\mathrm{5}\:{m} \\ $$$${i}.{e}.\:{the}\:{lineman}\:{can}'{t}\:{block}\:{the}\:{ball}. \\ $$
Commented by Abdoulaye last updated on 24/Feb/21
$$ \\ $$$${h}\mathrm{ow}\:\mathrm{do}\:\mathrm{you}\:\mathrm{do}\:\mathrm{these}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{figures}\: \\ $$$$\mathrm{on}\:\mathrm{this}\:\mathrm{app}? \\ $$
Commented by mr W last updated on 24/Feb/21
$${using}\:{the}\:{app}\:{Lekh}. \\ $$
Commented by mr W last updated on 25/Feb/21
$${the}\:{question}\:{could}\:{also}\:{mean}\:{this}: \\ $$
Commented by mr W last updated on 25/Feb/21
Commented by I want to learn more last updated on 25/Feb/21
$$\mathrm{Ohhh},\:\mathrm{alright}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$