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A particle is projected with a velocity of 10 ${ms}^{- 1}$ 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 = 10 ${ms}^{- 2}$ )

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zero

$10 {ms}^{- 2}$

$5 {ms}^{- 2}$

$2.5 {ms}^{- 2}$

answer is D.

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Detailed Solution

$For Particle Mg = \frac{M {(ucosθ)}^{2}}{R}$
$\Rightarrow R = \frac{u^{2} \cos^{2} θ}{g}$
$For Bird, a_{n} = \frac{v^{2}}{R} = \frac{v^{2}}{u^{2} \cos^{2} θ} \times g$
$= \frac{4^{2}}{10^{2} \cos^{2} 37^{0}} \times 10$ $= 2.5 {ms}^{- 2}$

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