A particle is projected from a point O with a velocity u at an angle $\alpha$ (upwards) to the horizontal. At a certain point P, it moves at right angles to its initial direction. It follows that

$\text{ (d) }\mathrm{vcos}(90-\mathrm{\alpha })=\mathrm{ucos}\mathrm{\alpha }$

                $\mathrm{v}=\mathrm{ucot}\mathrm{\alpha }$

$\text{ (c) }\tan (-(90-\mathrm{\alpha }\left)\right)=\frac{\mathrm{usin}\mathrm{\alpha }-\mathrm{gt}}{\mathrm{ucos}\mathrm{\alpha }}$

$\Rightarrow \mathrm{t}=\frac{\mathrm{u}}{\mathrm{gsin}\mathrm{\alpha }}$

$\text{ (b) }\mathbf{r}=(\mathrm{ucos}\mathrm{\alpha })\left(\frac{\mathrm{u}}{\mathrm{gsin}\mathrm{\alpha }}\right)\widehat{\mathbf{i}}+\left(\mathrm{usin}\mathrm{\alpha }\left(\frac{\mathrm{u}}{\mathrm{gsin}\mathrm{\alpha }}\right)\right)-\frac{1}{2}\mathrm{g}{\left(\frac{\mathrm{u}}{\mathrm{gsin}\mathrm{\alpha }}\right)}^2\widehat{\mathbf{j}}$

$\begin{array}{l}\mathbf{r}=\frac{{\mathrm{u}}^2}{\mathrm{g}}\cot \mathrm{\alpha }\widehat{\mathbf{i}}+\left(\frac{{\mathrm{u}}^2}{\mathrm{g}}-\frac{{\mathrm{u}}^2}{2\mathrm{g}}{\mathrm{cosec}}^2\mathrm{\alpha }\right)\widehat{\mathbf{j}}\\ \mathbf{r}=\frac{{\mathrm{u}}^2}{\mathrm{g}}\cot \mathrm{\alpha }\widehat{\mathbf{i}}+\frac{{\mathrm{u}}^2}{\mathrm{g}}\left(1-\frac{{\mathrm{cosec}}^2\mathrm{\alpha }}{2}\right)\widehat{\mathbf{j}}\\ \mathrm{OP}=\sqrt{{\left(\frac{{\mathrm{u}}^2}{\mathrm{g}}\cot \mathrm{\alpha }\right)}^2+{\left(\frac{{\mathrm{u}}^2}{\mathrm{g}}\left(1-\frac{{\mathrm{cosec}}^2\mathrm{\alpha }}{2}\right)\right)}^2}\end{array}$

$\text{ (a) }\tan \mathrm{\theta }=\frac{\frac{{\mathrm{u}}^2}{\mathrm{g}}\left(1-\frac{{\mathrm{cosec}}^2\mathrm{\alpha }}{2}\right)}{\frac{{\mathrm{u}}^2}{2}\cot \mathrm{\alpha }}$