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Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A particle is projected with velocity $20 m s^{- 1}$ 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
( $g = 10 m s^{- 1}$ )

Options:

Answer:

Solution:

$V c o s 30 ° = u c o s 60 ° \Rightarrow V = \frac{20}{\sqrt{3}} m s^{- 1}$
$a_{c} = g c o s 30 ° = 10 \times \frac{\sqrt{3}}{2}$
$∴ r = \frac{V^{2}}{a_{c}} = \frac{80}{3 \sqrt{3}} m$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Circular MotionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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