A partical when projected in vertical plane moves along smooth surface with initial velocity `20 ms^(-1)` at an angle of `60^(@)` , so that its normal reation on the surface remain zero throughout the motion. Then the slope of the surface at height `5 m` from the point of projection will be

Options

A. `sqrt3`
B. `1`
C. `2`
D. None of these

Answer

D

Explanation

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`.