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

Let ‘t’ is time the two stones meet for the dropped stone S=Ut+12at2by question, S=H3,a=+g,U=0 H3=0+12gt2---(1) ⇒t=2H3g---(2)  For projected stone  a=−g,  S=2H3 From  S=Ut+12at2  substitute given values in this equation,         ∴2H3=Ut−12gt2---(3) add equations (1) and (3) H=Ut ⇒H2=U2t2 substitute equation (2) ⇒H2=U22H3g acceleration due to gravity=g=10m/s2; H=85m U=3gH2=35.7 m/s=36m/s