A projectile is fired on a horizontal ground. Variation of slope of its trajectory with horizontal displacement is shown in figure. (Take, g =10m/ s² ). Choose the correct statement(s).

$\begin{array}{l}\mathrm{y}=\mathrm{xtan}\mathrm{\theta }-\frac{{\mathrm{gx}}^2}{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}\\ \text{ Slope }=\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \mathrm{\theta }-\frac{\mathrm{gx}}{{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}\\ \text{ At }\mathrm{x}=0;\text{ Slope }=\tan \mathrm{\theta }=1\\ \mathrm{\theta }={45}^{\circ }\\ \text{ At }\mathrm{x}=10;\text{ Slope }=1-\frac{\mathrm{g}\left(10\right)}{{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}=0\\ {\mathrm{u}}^2{\cos }^2\mathrm{\theta }=100\\ \Rightarrow \mathrm{ucos}\mathrm{\theta }=10\\ \Rightarrow \mathrm{u}=10\sqrt{2}\mathrm{m}/\mathrm{s}\\ \mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}=20\mathrm{m}\\ \mathrm{T}=\frac{2\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}=2\mathrm{s}\\ \int \left(\text{ slope }\right)\mathrm{dx}=\int \mathrm{dy}=\text{ vertical displacement }\end{array}$