AP EAPCET 19-Aug-2021 Shift 1 Paper

According to the question,
Time taken by ball to reach on floor when lift is stationary is t1 and time taken by ball to reach on floor when lift is moving up with constant acceleration a is t2.
Since, s=ut+

1

2

at2
Therefore, in case 1
s=0t1+

1

2

gt12 [∵a=g]
⇒s=

1

2

gt12 ...(i)
If anet is net acceleration of ball, then in case 2
when lift is going up with acceleration a
aDCt=(g+a)
⇒s=

1

2

(g+a)t22
From Eqs. (i) and (ii), we get

1

2

gt12=

1

2

(g+a)t22
∴t1>t2