Question Number 103421 by aurpeyz last updated on 15/Jul/20
Answered by Worm_Tail last updated on 15/Jul/20
$${v}_{{y}} ^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\mathrm{2}{gs}_{{y}} \Rightarrow\mathrm{3}.\mathrm{7}^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\left(\mathrm{2}×\mathrm{10}×\mathrm{7}.\mathrm{5}\right) \\ $$$$\mathrm{13}.\mathrm{69}+\mathrm{150}=\left({usin}\theta\right)^{\mathrm{2}} \Rightarrow\left({usin}\theta\right)^{\mathrm{2}} =\mathrm{163}.\mathrm{69} \\ $$$${h}_{{max}} =\frac{\left({usin}\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\mathrm{163}.\mathrm{69}}{\mathrm{20}}=\mathrm{8}.\mathrm{1845}{m} \\ $$$${T}=\frac{\mathrm{2}\left({usin}\theta\right)}{{g}}=\frac{\mathrm{2}×\sqrt{\mathrm{163}.\mathrm{69}}}{\mathrm{10}}=\mathrm{2}.\mathrm{5588}{s} \\ $$$${R}={v}_{{x}} {T}=\mathrm{10}.\mathrm{4}×\mathrm{2}.\mathrm{5588}=\mathrm{26}.\mathrm{61}{m} \\ $$
Commented by aurpeyz last updated on 15/Jul/20
$$\mathrm{thanks}\:\mathrm{sir} \\ $$