EAMCET Engineering 2008 Solved Paper

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+‌

t

√2