A ball is projected horizontally on the inclined plane as shown in the figure. If time of flight is 5 s. Find the range of projectile on inclined plane. 

Here, ${\mathrm{u}}_{\mathrm{x}}={\mathrm{v}}_0\cos {37}^{\circ },{\mathrm{u}}_{\mathrm{y}}={\mathrm{v}}_0\sin {37}^{\circ }$

$\begin{array}{l}{\mathrm{a}}_{\mathrm{x}}=\mathrm{gsin}{37}^{\circ }=10\times \frac{3}{5}=6{\mathrm{ms}}^{-2}\\ \text{ and }{\mathrm{a}}_{\mathrm{y}}=-\mathrm{gcos}{37}^{\circ }=-10\times \frac{4}{5}=-8{\mathrm{ms}}^{-2}\\ \because \mathrm{y}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2\\ \Rightarrow 0=\left({\mathrm{v}}_0\sin {37}^{\circ }\right)\mathrm{T}-\frac{1}{2}\times 8\times {\mathrm{T}}^2\\ \therefore \mathrm{T}=\frac{2{\mathrm{v}}_0\sin {37}^{\circ }}{8}\text{ or }5=\frac{2\times {\mathrm{v}}_0}{8}\times \frac{3}{5}\\ \therefore {\mathrm{v}}_0=\frac{100}{3}{\mathrm{ms}}^{-1}\\ \therefore \mathrm{R}={\mathrm{u}}_{\mathrm{x}}\mathrm{T}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{T}}^2=\frac{100}{3}\times \frac{4}{3}\times 5+\frac{1}{2}\times 6\times 25\\ =\frac{2000}{9}+75\\ \\ =297.22m\end{array}$