TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

In last 1 s, ball covers H/2 distance. Let total time for falling by H height is t second.
So, ‌‌H=‌

1

2

gt2....(i)
In (t−1) second, ball falls by ‌

H

2

height.
So, ‌‌‌

H

2

=‌

1

2

g(t−1)2...(ii)
Now, dividing Eq. (i) by Eq. (ii), we get
‌

H

H/2

=‌

‌ 1 2 gt2

1 2 g(t−1)2

⇒√2=‌

t

t−1

√2t−√2=t
t=‌

√2

√2−1

=‌

√2(√2+1)

(√2−1)(√2+1)

=2+√2=2+1.41
⇒‌‌t=3.41s