The correct option is A 40 m
Let the maximum height attained by particle be H
Taking vertically upward direction as +ve and applying kinematic equation,
v2=u2+2as ...(i)
where at maximum height v=0,
Displacement of particle from half the maximum height to maximum height is s=+H2
& velocity of particle at distance equal to half of maximum height is u=+20 m/s
Also, a=−g=−10 m/s2
Putting all values in Eq (i),
(0)2=(20)2−2×10×(H2)
or 400=10H
∴H=40 m