A palne is inclined at an angle `alpha=30^(@)` with respect to the horizontal. A particle is projected with a speed `u=2ms^(-1)` from the base of the plane, as shown in figure. The distance (in metre) from the base, at which the particle hits the plane is close to: (Take `g=10ms^(-2)`)

Explanation

On an inclined plane, time fo flight (T) is given by
`T=(2usin theta)/(g cos alpha)`
Substituting the value we get
`T=((2)(2sin 15^(@)))/(g cos 30^(@))=(4 sin 15^(@))/(10 cos 30^(@))`
Distance `S=(2 cos 15^(@))T-1/2g sin 30^(@)(T)^(2)`

`=(2cos 15^(@))4/10(sin 15^(@))/(10 cos 30^(@))-(1/2 xx 10 sin 30^(@))(16sin^(2)15^(@))/(100 cos^(2)30^(@))`
`=(16sqrt(3)-16)/60=0.195m`