Question Number 176146 by peter frank last updated on 13/Sep/22
Answered by mr W last updated on 13/Sep/22
Commented by mr W last updated on 13/Sep/22
$${b}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{t}=\sqrt{\frac{\mathrm{2}{b}}{{g}}} \\ $$$${u}_{{x}} =\frac{{a}}{\mathrm{2}{t}}=\frac{{a}}{\mathrm{2}}\sqrt{\frac{{g}}{\mathrm{2}{b}}} \\ $$$${u}_{{y}} ={gt}=\sqrt{\mathrm{2}{bg}} \\ $$$$ \\ $$$${y}={kx}\left({a}−{x}\right) \\ $$$${b}={k}×\frac{{a}}{\mathrm{2}}×\left({a}−\frac{{a}}{\mathrm{2}}\right)\:\Rightarrow{k}=\frac{\mathrm{4}{b}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{y}=\frac{\mathrm{4}{bx}\left({a}−{x}\right)}{{a}^{\mathrm{2}} } \\ $$
Commented by peter frank last updated on 14/Sep/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$