A body is projected at t=0 with a velocity 10 ~ms ⁻¹ at an angle of 60^ with the horizontal. The radius of curvature of its trajectory at t=1 s is R . Neglecting air resistance and taking acceleration due to gravity g =10 ~ms ⁻² , the value of R is:

Step-by-step solution

Horizontal component of velocity v _ x =10 60^ =5 ~m / s vertical component of velocity v_ y =10 30^ =5 3 ~m / s After t =1 sec Horizontal component of velocity v _ x =5 ~m / s Vertical component of velocity v_ y =|(5 3 -10)| m / s =10-5 3 Centripetal, acceleration a _ n = v ² R R = v _ x ²+ v _ y ² a _ n = 25+100+75-100 3 10 From figure (using (i)) = 10-5 3 5 =2- 3 =15^ R = 100(2- 3 ) 10 15 =2.8 ~m