The ball is thrown vertically upwards then according to equation of motion (0)2−u2=−2gh . . . (i) and 0=u−gt . . . (ii) From Eqs. (i) and (ii), h=‌
gt2
2
When the ball is falling downwards after reaching the maximum height s‌=ut′+‌
1
2
g(t′)2 ‌
h
2
‌=(0)t′+‌
1
2
g(t′)2 ⇒‌‌t′‌=√‌
h
g
t′‌=‌
t
√2
Hence, the total time from the time of projection to reach a point at half of its maximum height while returning =t+t′ =t+‌