A particle is projected from an inclined plane `OP_1` from A with velocity `v_1 = 8ms^(-1)` at an angle `60^@` with horizontal. An another particle is projected at the same instant from B with velocity `v_2 = 16 ms^(-1)` and perpendicular to the plane `OP_2` as shown in figure. After time `10(sqrt3)` s there separation was minimum and found to be 70m. Then find distance AB.

Options

A. 250m
B. 500m
C. 750m
D. 1000m

Answer

a

Explanation

`(v_(B.A))_("horizontal")=v_(2)cos60+v_(1)cos 60=12m//s`.
`(V_(B.A.))_("vertical")=v_(2)sin 60-v_(1) sin 60=4sqrt(3)m//s`
`v_(B.A.)=sqrt((12)^(2)+(4sqrt(3)^(2))=sqrt(192)`
`x=v_(21)t=sqrt(192)xx10sqrt(3)=240m`

From `DeltaABC`
`AB^(2)=AC^(2)+BC^(2)`
`AB=250m`