A wet open umbrella is held vertical and it whirld about the handle at a uniform rate of 21 revolutions in 44 s . If the rim of the umbrella is circle of 1 m in diameter and the height of the rim above the flour is 4.9 m , the locus of the drop is a circle of radius

Step-by-step solution

From equation of motion h=u t+ 1 2 g t^2 where u is initial velocity and t is time. Since, u=0 t= 2 h g = 2 4.9 9.8 =1 ~s The horizontal range of the drop =x , then x= ( v_t o ) t Also, = t = 21 2 44 =3 rad / s Tangential speed 0.5 3 = 1.5x=1.5 1=1.5 ~m aligned Locus of drop = x^2+r^2 & = (1.5)^2+(0.5)^2 & = 2.5 ~m aligned