The length of inclined plane is 15 m. A stone is projected with speed v₀ from point A to cross through point B. If minimum value of v₀ is 3x ms^-1. The value of x is _______.

For minimum value of v₀ the maximum range on inclined plane should be AB - 15m.

$\text{ The maximum range will be }{R}_{max}=\frac{v_0^2}{g\left(1+\sin {30}^{\circ }\right)}$

$\begin{array}{l}\text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}_0^2=15\times 10\times \frac{3}{2}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}_0^2={15}^2\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}_0=15{\mathrm{ms}}^{-1}=3x\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=5\end{array}$