Correct Answer - D
It implies that the given surface is the path of the given projectile
`y = x tan theta - (gx^(2))/(2 u^(2) cos^(2) theta)`
`= x tan 60^(@) - ((10)x^(2))/((2) (20)^(2) cos^(2) 60^(@))`
`y = sqrt3 x - 0.05 x^(2)` …(i)
Slope, `(dy)/(dx) = sqrt(3) - 0.1 x` ...(ii)
At `y = 5 m`
`5 = sqrt3x - 0.05x^(2)`
or `0.05 - sqrt(3)x + 5 = 0`
`x = (sqrt3 pm sqrt(3 - 1))/(0.1) = (sqrt3 pm sqrt2)/(0.1)`
From Eq (ii) slope at these two points are `- sqrt2` and `sqrt2`.