A particle when projected in vertical plane moves along smooth surface with initial velocity $20{ \mathrm{ms}}^{-1}$ at an angle of $60°,{80}^o$ that its normal reaction on the surface remains zero throughout the motion. Then, the slope of the surface at height $5 \mathrm{m}$ from the point of projection will be 

It implies that the given surfice is the path of the given projectile

$$\begin{gathered} y=x \tan \theta -\frac{{gx}^2}{2a^2{\cos }^2\theta } \\ =x \tan {60}^2-\frac{\left(10\right)x^2}{\left(2\right)(20{)}^2{\cos }^2{60}^2} \end{gathered}$$ 

       $y=\sqrt{3}x-0.05x^2$                       …………………(i)

$\text{ Slope, }\frac{dy}{dx}=\sqrt{3}-0.1x$            ………………….(ii)

$\text{ at }y=5m$

 $$\begin{gathered} 5=\sqrt{3}x-0.05x^2 \\ 0.05x^2-\sqrt{3}x+5=0 \\ x=\frac{\sqrt{3}\pm \sqrt{3-1}}{0.1}=\frac{\sqrt{3}\pm \sqrt{2}}{0.1} \end{gathered}$$

From Eq. (ii) slope at these two points are, $-\sqrt{2}$ and $\sqrt{2}$.