A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t1 and t2 seconds respectively, then
If a stone is dropped from height h
then h=12gt2 ....(i)
If a stone is thrown upward with velocity u then,
h=−ut1+12gt21 ....(ii)
If a stone is thrown downward with velocity u then
h=ut2+12gt22 ...(iii)
From (i) (ii) and (iii) we get
−ut1+12gt21=12gt2 ..(iv)
ut2+12gt22=12gt2 ..(v)
Dividing (iv) and (v) we get
∴−ut1ut2=12g(t2−t21)12g(t2−t22)
or −t1t2=t2−t21t2−t22
By solving t=√t1t2