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 )

Question

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 ) (A) 60√3 m (B) (80/√3) m (C) 40√3 m (D) (80/3 √3) m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-qz6yrqqhm9ijlfxo.jpg" /> Vcos30° =ucos60° ⇒ V=(20/√3) ms- 1 ac=gcos30° =10× (√3/2) ∴ r=(V2/ac)=(80/3 √3) m

Q. A particle is projected with velocity $20 m s^{- 1}$ at angle $6 0^{\circ}$ with horizontal. The radius of curvature of trajectory, at the instant when velocity of projectile becomes perpendicular to velocity of projection, is
( $g = 10 m s^{- 1}$ )

$603 m$

$\frac{80}{3} m$

$403 m$

$\frac{80}{3 3} m$

$V cos 3 0^{\circ} = u cos 6 0^{\circ} \Rightarrow V = \frac{20}{3} m s^{- 1}$
$a_{c} = g cos 3 0^{\circ} = 10 \times \frac{3}{2}$
$∴ r = \frac{V ^{2}}{a _{c}} = \frac{80}{3 3} m$

Q. 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 )

603​m

3​80​m

403​m

33​80​m