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