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

Answer

A

Explanation

`(V_(B.A))_("horizontal")=v_(2) cos 60+v_(1) cos 60=12m//s`
`(V_(B.A))_("vertical")=v_(2) sin 60-v_(1) sin 60=4sqrt3m//s`
`V_(B.A)=sqrt((12)^(2)+(4sqrt3))=sqrt192`
`x=v_(21)t=sqrt192xx10sqrt3=240m`
from `Delta l e ABC`
`AB^(2)=Ac^(2)+Bc^(2)`
`AB=250`