A stone dropped from a building of height h reaches the ground after t seconds. From the same building if two stones are thrown (one upwards and the other downwards) with the same velocityu and they reach the ground after t1 and t2 seconds respectively, then the time interval t is

For the stone dropped with zero initial velocity. We have h=12gt2 …………………(i) For the stone thrown upwards with velocity u, we haveh=−ut1+12gt12. ⇒ut1+12gt12−−−−−−−(ii)For the stone thrown downwards with velocity u, we haveh=ut2+12gt22. ⇒ut1+12gt22−−−−−−−(iii)Notice that the displacement of the stone in the three case is the same, equal to h. Using (i) to (ii) and (iii) we have12gt2=−ut1+12gt12⇒ut1=12gt12−12gt212gt2=ut2+12gt22⇒ut2=12gt2−12gt22Dividing the two equations ,we havet1t2=t12−t2t2−t22Which gives t=t1t2. Hence the correct choice is (c).