For a particle projected from the ground with speed u=20 ms- 1 at an angle of 60° with horizontal. The radius of curvature of the path of the particle, when its velocity makes an angle of 30° with horizontal will be.... (.g =10 ms- 2.) .

Question

For a particle projected from the ground with speed u=20 ms- 1 at an angle of 60° with horizontal. The radius of curvature of the path of the particle, when its velocity makes an angle of 30° with horizontal will be.... (.g =10 ms- 2.) . (A) 10.6 m (B) 12.8 m (C) 15.4 m (D) 24.2 m

Tardigrade Admin · Accepted Answer

Let v be the velocity of particle when it makes

3 0^{ο}

with horizontal
Solution

v cos 30 = u cos 60

v \frac{3}{2} = (20) (\frac{1}{2})

v = \frac{20}{3} m s^{- 1}

now use centripetal acceleration

g cos 3 0^{ο}

m g cos 3 0^{ο} = \frac{mv ^{2}}{R}

R = \frac{( v ) ^{2}}{g cos ( 30 ) ^{ο}} = \frac{( \frac{20}{3} ) ^{2}}{10 ( \frac{3}{2} )}

= 15.4 m

Q. For a particle projected from the ground with speed u=20ms−1 at an angle of 60∘ with horizontal. The radius of curvature of the path of the particle, when its velocity makes an angle of 30∘ with horizontal will be.... (g=10ms−2) .

10.6 m

12.8 m

15.4 m

24.2 m

Let v be the velocity of particle when it makes 30ο with horizontal vcos30=ucos60 v23​​=(20)(21​) v=3​20​ms−1 now use centripetal acceleration gcos30ο mgcos30ο=Rmv2​ R=gcos(30)ο(v)2​=10(23​​)(3​20​)2​ =15.4m

Q. For a particle projected from the ground with speed $u = 20 m s^{- 1}$ at an angle of $6 0^{\circ}$ with horizontal. The radius of curvature of the path of the particle, when its velocity makes an angle of $3 0^{\circ}$ with horizontal will be.... $(g = 10 m s^{- 2})$ .