A body is project at t= 0 with a velocity `10 ms^-1` at an angle of `60 ^(@)` with the horizontal .The readius of curvature of its trajectory at t=1s is R. Neglecting air resitance and taking acceleration due to gravity `g= 10 ms^-2`, the value of R is :

Explanation

Horizontal component of velocity
`v_(x)=10 cos 60^(@)=5m//s`
vertical component of velocity
`v_(y)=10 cos 30^(@)=5sqrt(3)m//s`
After t=1 sec
Horizontal component of velocity `v_(x)=5m//s`
Vertical component of velocity
`v_(y)=|(5sqrt(3)-10)|m//s=10-5sqrt(3)`
Centripetal, acceleration `a_(n)=(v^(2))/R`
`impliesR=(v_(x)^(2)+v_(y)^(2))/(a_(n))=(25+100+75-100sqrt(3))/(10 cos theta) `...............i
From figure (using (i))
`tan theta =(10 -5sqrt(3))/5=2-sqrt(3)impliestheta=15^(@)`
`R=(100(2-sqrt(3)))/(10 cos 15) =2.8m`