JEE Main 25 Feb 2021 Shift 2 Solved Paper

Let the total height of building be x. TA‌‌=5m
TB‌‌=25m
∴‌‌AG=x−5 and BC=x−25
For initial conditions, from second equation of motion under gravity,
s=ut+1∕2gt2
where, g=10ms−2
∴‌‌5=0+1∕2×10t2⇒t=1s
Now, by first equation of motion under gravity,
vA‌‌=u+gt
‌‌=0+10=10ms−1
From second equation of motion,
x−5=vAt+1∕2gt2 . . . (i)
Similarly, x−25=1∕2gt2
Put the above value in Eq. (i), we get
x−5‌‌=10t+x−25
20‌‌=10t⇒t=2s
Put the value of t in Eq. (i), we get
x−5‌‌=10×2+1∕2×10×4
x−5‌‌=20+20
x‌‌=45m