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Question Number 195874 by Humble last updated on 12/Aug/23

Answered by mr W last updated on 12/Aug/23

x=v_0 sin θ t+((g cos α)/2) t^2 z=v_0 cos θt−((g sin α)/2) t^2 a) z_(max) =(((v_0 cos θ)^2 )/(2g sin α))=(((4×((√3)/2))^2 )/(2×9.8×sin 45°))=0.87 m b) z=v_0 cos θt−((g sin α)/2) t^2 =0 ⇒t=((2v_0 cos θ)/(g sin α)) x_(max) =(v_0 sin θ+((g cos α)/2)×((2v_0 cos θ)/(g sin α)))((2v_0 cos θ)/(g sin α)) =(sin θ+((cos θ)/(tan α)))((2v_0 ^2 cos θ)/(g sin α)) =((1/2)+((√3)/2))((2×4^2 ×(√3)×(√2))/(2×9.8)) =5.46 m

$${x}={v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}+\frac{{g}\:\mathrm{cos}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} \\ $$$${z}={v}_{\mathrm{0}} \mathrm{cos}\:\theta{t}−\frac{{g}\:\mathrm{sin}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} \\ $$$$\left.{a}\right) \\ $$$${z}_{{max}} =\frac{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}\:\mathrm{sin}\:\alpha}=\frac{\left(\mathrm{4}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}.\mathrm{8}×\mathrm{sin}\:\mathrm{45}°}=\mathrm{0}.\mathrm{87}\:{m} \\ $$$$\left.{b}\right) \\ $$$${z}={v}_{\mathrm{0}} \mathrm{cos}\:\theta{t}−\frac{{g}\:\mathrm{sin}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{cos}\:\theta}{{g}\:\mathrm{sin}\:\alpha} \\ $$$${x}_{{max}} =\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta+\frac{{g}\:\mathrm{cos}\:\alpha}{\mathrm{2}}×\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{cos}\:\theta}{{g}\:\mathrm{sin}\:\alpha}\right)\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{cos}\:\theta}{{g}\:\mathrm{sin}\:\alpha} \\ $$$$\:\:=\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\alpha}\right)\frac{\mathrm{2}{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}\:\theta}{{g}\:\mathrm{sin}\:\alpha} \\ $$$$\:\:=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\frac{\mathrm{2}×\mathrm{4}^{\mathrm{2}} ×\sqrt{\mathrm{3}}×\sqrt{\mathrm{2}}}{\mathrm{2}×\mathrm{9}.\mathrm{8}} \\ $$$$\:\:=\mathrm{5}.\mathrm{46}\:{m} \\ $$

Commented by Humble last updated on 12/Aug/23

Great! Thanks sir.

$$\mathrm{Great}! \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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