A stone is dropped from a top of a tall tower of height H=85m. Another stone is projected up from the bottom  of tower at the same instant with speed ‘U’. If  they meet at height 2H3  from ground then  the value ‘U’ nearly is (g=10m/s2 ) (round off after decimal)

Let ‘t’ is time the two stones meet for the dropped stone  S=Ut+12at2by question, displacement=S=H3; acceleration due to gravity=a=+g;  initial velocity=U=0  H3=12gt2---(1) ⇒t=2H3g---(2)   For projected stone a=−g,  S=2H3 S=Ut+12at2          ∴2H3=Ut−12gt2---(3) substitute equation (1) in (3)  H=Ut ⇒H2=U2t2 substituting eqn (2) ⇒H2=U22h3g   U=3gH2=35.7 m/s   = 36m/sinitial velocity of projected body