Question-155294

Question Number 155294 by peter frank last updated on 28/Sep/21

Answered by peter frank last updated on 30/Sep/21

Commented by peter frank last updated on 30/Sep/21

Point P=point B R_1 =((v_o ^2 sin 2θ_1 )/g) =0.899m θ_1 =46 R_2 =((v_o ^2 sin 2θ_2 )/g) =0.899m θ_2 =44 R_1 =R_2 hencd both particle focused on the same point P (b) h=h_2 −h_1 h_2 =((v_o ^2 sin^2 θ_2 )/(2g)) h_1 =((v_o ^2 sin^2 θ_1 )/(2g)) g=a=4×10^(13) m/s^2 hence h=0.233m

$$\mathrm{Point}\:\mathrm{P}=\mathrm{point}\:\mathrm{B}\: \\ $$$$\mathrm{R}_{\mathrm{1}} =\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta_{\mathrm{1}} }{\mathrm{g}}\:=\mathrm{0}.\mathrm{899m}\:\:\:\theta_{\mathrm{1}} =\mathrm{46} \\ $$$$\mathrm{R}_{\mathrm{2}} =\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta_{\mathrm{2}} \:}{\mathrm{g}}\:=\mathrm{0}.\mathrm{899m}\:\:\:\theta_{\mathrm{2}} =\mathrm{44} \\ $$$$\mathrm{R}_{\mathrm{1}} =\mathrm{R}_{\mathrm{2}} \:\:\mathrm{hencd}\:\mathrm{both}\:\mathrm{particle}\: \\ $$$$\mathrm{focused}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same}\:\mathrm{point}\:\mathrm{P} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{h}=\mathrm{h}_{\mathrm{2}} −\mathrm{h}_{\mathrm{1}} \\ $$$$\mathrm{h}_{\mathrm{2}} =\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta_{\mathrm{2}} \:}{\mathrm{2g}} \\ $$$$\mathrm{h}_{\mathrm{1}} =\frac{\mathrm{v}_{\mathrm{o}} ^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta_{\mathrm{1}} \:}{\mathrm{2g}} \\ $$$$\mathrm{g}=\mathrm{a}=\mathrm{4}×\mathrm{10}^{\mathrm{13}} \mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\mathrm{hence} \\ $$$$\mathrm{h}=\mathrm{0}.\mathrm{233m} \\ $$

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Question-155294

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