Equation of trajectory of a particle moving in xy-plane is given by $y=10x-2x^2$  (x and y are in meters). Assume acceleration of gravity as 10m/s² in negative y-direction. Choose correct statement(s).

$\begin{array}{l}\mathrm{y}=10\mathrm{x}-2{\mathrm{x}}^2=10\mathrm{x}\left(1-\frac{\mathrm{x}}{5}\right)\\ \text{ Comparing it with }\mathrm{y}=\mathrm{xtan}\mathrm{\theta }\left(1-\frac{\mathrm{x}}{\mathrm{R}}\right)\\ \mathrm{R}=5\mathrm{m}\text{ and }\tan \mathrm{\theta }=10 \Rightarrow \mathrm{\theta }=84.3^0\\ \frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}=5\\ \Rightarrow {\mathrm{u}}^{2}2\sin \mathrm{\theta cos}\mathrm{\theta }=50\\ \Rightarrow 2{\mathrm{u}}^2\frac{10}{\sqrt{101}}\frac{1}{\sqrt{101}}=50\\ \Rightarrow {\mathrm{u}}^2=25\times 10.1\end{array}$
$\begin{array}{l}\Rightarrow \mathrm{u}=5\sqrt{10.1}\mathrm{m}/\mathrm{s}\\ \mathrm{H}=\frac{{\mathrm{u}}^2{\sin }^2\theta }{2\mathrm{g}}=\frac{25\left(10.1\right)}{20}\frac{100}{101}=12.5\mathrm{m}\\ \tan \varphi =\frac{\mathrm{usin}\theta -\mathrm{gt}}{\mathrm{ucos}\theta }\\ \tan \varphi =\frac{\mathrm{usin}\theta -\mathrm{g}\left(\frac{\mathrm{x}}{\mathrm{ucos}\theta }\right)}{\mathrm{ucos}\theta }\\ At \varphi ={45}^o\\ \Rightarrow \frac{{\mathrm{u}}^2\sin \theta \cos \theta -\mathrm{gx}}{{\mathrm{u}}^2{\cos }^2\theta }=1\Rightarrow \mathrm{x}=2.25m\end{array}$