1)

 A bird is sitting on the top of vertical pole 20m high and its elevation from a point O on the ground 45° . It flies off horizontally straight away from the point O. After 1s,  the elevatiion  of the bird from O is reduced to 30° . Then the speed (in m/s)  of the bird is 


A) 40(31)

B) 202

C) 40(32)

D) 20(31)

Answer:

Option D

Explanation:

In  OA1B1,

    tan450=A1B1OB120OB1=1

        OB1=20

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OA2B2,

   tan300=20OB2

            OB2=203

                     B1B2+OB1=203

                     B1B2=20320            

                B1B2=20(31)m

 Now, speed = DistanceTime=20(31)1

                              =   20(31)  m/s