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Question-193336




Question Number 193336 by Abdulz last updated on 10/Jun/23
Commented by Abdulz last updated on 10/Jun/23
please, help
$${please},\:{help} \\ $$
Answered by Humble last updated on 11/Jun/23
  α=45°  ,    θ=30°  v_0 =4m/s  v_2 =? g=9.8m/s  H_(max) =(u^2 /(2a))=((v_0 ^2 cos^2 (θ))/(2gsin(α)))  = ((4^2 cos^2 (30))/(2×9.8×sin(45)))  =((16×(((√3)/2))^2 )/(19.6×((√2)/2))) =((16×(3/4))/((19.6(√2))/2))   =16×(3/4)×(2/(19.6(√2)))  =16×(3/2)×(1/(19.6(√2)))  =((24)/(19.6(√2))) =((30(√2))/(49))  =0.87m  (b) R=((2v_0 ^2 cos(θ)sin(θ))/g)  R=((2(4)^2 cos(30)sin(30))/(9.8))  R=((32×((√3)/2)×(1/2))/(9.8))  =(((32(√3))/4)/(9.8))  R=((32(√3))/4)×(1/(9.8))  =((32(√3))/(39.2))  R=((40(√3))/(49)) =1.4m
$$ \\ $$$$\alpha=\mathrm{45}°\:\:,\:\:\:\:\theta=\mathrm{30}°\:\:{v}_{\mathrm{0}} =\mathrm{4}{m}/{s} \\ $$$${v}_{\mathrm{2}} =?\:{g}=\mathrm{9}.\mathrm{8}{m}/{s} \\ $$$${H}_{{max}} =\frac{{u}^{\mathrm{2}} }{\mathrm{2}{a}}=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\theta\right)}{\mathrm{2}{gsin}\left(\alpha\right)} \\ $$$$=\:\frac{\mathrm{4}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{30}\right)}{\mathrm{2}×\mathrm{9}.\mathrm{8}×{sin}\left(\mathrm{45}\right)} \\ $$$$=\frac{\mathrm{16}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{19}.\mathrm{6}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\mathrm{16}×\frac{\mathrm{3}}{\mathrm{4}}}{\frac{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{2}}}\: \\ $$$$=\mathrm{16}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{2}}{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{16}×\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{24}}{\mathrm{19}.\mathrm{6}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{30}\sqrt{\mathrm{2}}}{\mathrm{49}}\:\:=\mathrm{0}.\mathrm{87}{m} \\ $$$$\left({b}\right)\:{R}=\frac{\mathrm{2}{v}_{\mathrm{0}} ^{\mathrm{2}} {cos}\left(\theta\right){sin}\left(\theta\right)}{{g}} \\ $$$${R}=\frac{\mathrm{2}\left(\mathrm{4}\right)^{\mathrm{2}} {cos}\left(\mathrm{30}\right){sin}\left(\mathrm{30}\right)}{\mathrm{9}.\mathrm{8}} \\ $$$${R}=\frac{\mathrm{32}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{9}.\mathrm{8}}\:\:=\frac{\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{4}}}{\mathrm{9}.\mathrm{8}} \\ $$$${R}=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{9}.\mathrm{8}}\:\:=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{39}.\mathrm{2}} \\ $$$${R}=\frac{\mathrm{40}\sqrt{\mathrm{3}}}{\mathrm{49}}\:=\mathrm{1}.\mathrm{4}{m} \\ $$

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