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A particle is projected from the ground ...

A particle is projected from the ground at an angle of `60^(@)` with horizontal at speed `u = 20 m//s.` The radius of curvature of the path of the particle, when its velocity. makes an angle of `30^(@)` with horizontal is :
`(g=10 m//s^(2)`

A

10.6 m

B

12.8 m

C

15.4 m

D

24.2 m

Text Solution

Verified by Experts

The correct Answer is:
A
`bar(V)_(B//P)=(3hat(i)+hat(j))-(7hat(i)-2hat(j))`
`vec(f)_(k)=-hatv_(B//P)=(4)/(5)hat(i)-(3)/(5)hat(j)`
friction will be opposite to relative velocity.
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