A particle when projected in vertical plane moves along smooth surface with initial velocity `20 ms^(-1)` at an angle of `60^(@)` , so that its normal reaction 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. `30^(@)`
B. `45^(@)`
C. `tan^(-1) 2`
D. `tan^(-1) sqrt(2)`

Answer

D

Explanation

(Tough) If the normal reaction is always zero, the particle always moves under action gravitational force. Hence the path of particle (and also the nature of surface) is parabolic.

At a height h = 5 metre above the projection the speed of the particle is
`v_(2) = sqrt(u^(2) - 2gh) = sqrt(400 - 2 xx 10 xx 5)`
`= 10 sqrt(3) m//s`
The horizontal component of velocity does not change
`:. v cos theta = 20 cos 60^(@)`
`cos theta = (1)/(sqrt(2))` or `tan theta = sqrt(2)`
`rArr theta = tan_(1) sqrt(2)`