In the given figure , the angle of inclination of the inclined plane is `30^(@)`. A particle is projected with horizontal velocity `v_(0)` from height H. Find the horizotnal velocity `v_(0)` (in m/s) so that the particle hits the inclined plane perpendicu lar .Given H=4m, g`=10m//s^(2)`.

Answer

4

Explanation

`0=v_(0) cos 30^(@)-g sin 30^(@)t`
`rArr t=(v_(0)cos 30^(@))/(g sin 30^(@))` .....(i)
`-Hcos 30^(@)=-v_(0)sin30^(@)t-(1)/(2)g cos 30^(@)t^(2)` .......(ii)
By Eqns. (i) and (ii), we get
`H=(v_(0)^(2))/(g)[1+(cot^(2)alpha)/(2)]`
`rArr v_(0)=sqrt((2gH)/(5))=4m//s (alpha=30^(@))`