A particle is projected wityh velocity 20 `ms^(-1)` at angle `60^(@)` above horizontal up on an inclined plane having inclination `30^(@)`. Find its shortest distance from the inclined plane when it is moving `53^(@)` above horizontal.
(Using `=10ms^(-2)`)

Options

A. 11.1m
B. 8.2m
C. 3.3m
D. 14.4m

Explanation

`V_("vertical") U_("vertical")-"gt"`
`10 tan 53^(@)=20 sin 60^(@)-10t`
`(4)/(3)=sqrt(3)-1`
`t=sqrt(3)-(4)/(3)`
`s_(y)=u_(y)+a_(y)t^(2)`
`=10xx((3sqrt(3)-4)/(3))-(1)/(2)g cos 30^(@) ((3sqrt(3)-4)/(3))^(2)`
`=10xx((3sqrt(3)-4)/(3))-(1)/(2)xx10sqrt(3)/(2)((3sqrt(3)-4)/(3))^(2)`
`=3.99-2.5xx1.732((3xx1.732-4)/(7))^(2)`
`=3.99-4.33xx0.16`
`=3.99-0.688`
`=3.99-0.69=3.30`