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Q.

A particle is projected with a velocity of 10ms-1 at an angle 37° with horizontal. A bird moves along the path of projected particle with constant speed of 4 ms-1. Find the acceleration of bird when it is at highest point of path of particle.(g = 10ms-2)

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a

zero

b

10 ms-2

c

5 ms-2

d

2.5 ms-2

answer is D.

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Detailed Solution

For Particle Mg=Mucosθ2R⇒R=u2cos2θgFor Bird, an=v2R=v2u2cos2θ×g=42102cos2370×10= 2.5 ms-2

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