Question Number 103419 by aurpeyz last updated on 15/Jul/20
Answered by Worm_Tail last updated on 15/Jul/20
$${v}_{{y}} ={usin}\theta−{gt}\Rightarrow\mathrm{4}.\mathrm{8}={usin}\theta−\left(\mathrm{10}×\mathrm{0}.\mathrm{8}\right) \\ $$$$\mathrm{4}.\mathrm{8}+\mathrm{8}={usin}\theta\Rightarrow{usin}\theta=\mathrm{12}.\mathrm{8} \\ $$$${h}_{{max}} =\frac{\left({usin}\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\left(\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} }{\mathrm{20}}=\mathrm{8}.\mathrm{192}{m} \\ $$$$ \\ $$$${T}=\frac{\mathrm{2}\left({usin}\theta\right)}{{g}}=\frac{\mathrm{2}×\mathrm{12}.\mathrm{8}}{\mathrm{10}}=\mathrm{2}.\mathrm{56}{s} \\ $$$${R}={v}_{{x}} {T}=\mathrm{15}.\mathrm{5}×\mathrm{2}.\mathrm{56}=\mathrm{39}.\mathrm{68}{m} \\ $$
Commented by aurpeyz last updated on 15/Jul/20
$$\mathrm{thanks}\:\mathrm{sir} \\ $$