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  4. A particle is projected with velocity 20ms- 1 at angle 60° with horizontal. The radius of curvature of trajectory, at the instant when velocity of projectile becomes perpendicular to velocity of projection, is ( g=10ms- 1 )

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Q. A particle is projected with velocity $20ms^{- 1}$ at angle $60^\circ $ with horizontal. The radius of curvature of trajectory, at the instant when velocity of projectile becomes perpendicular to velocity of projection, is
( $g=10ms^{- 1}$ )

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
$Vcos30^\circ =ucos60^\circ \Rightarrow \, \, V=\frac{20}{\sqrt{3}} \, ms^{- 1}$
$a_{c}=gcos30^\circ =10\times \frac{\sqrt{3}}{2}$
$\therefore \, r=\frac{V^{2}}{a_{c}}=\frac{80}{3 \sqrt{3}} \, m$