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A particle is projected from an inclined...

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.

Text Solution

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The correct Answer is:
B
`|v_(21x)| = (v_1+v_2)cos 60^@ = 12m//s`
`|v_(21y)| = (v_2-v_1) sin 60^@ = 4(sqrt3) m//s`
`:. v_(21) = sqrt((12)^2+(4sqrt(3))^2) = (sqrt(192)) m//s`

`BC = (v_(21))t = 240m`
`AC = 70m`
Hence, `AB = sqrt((240)^2 + (70)^2) = 250m`.
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