A ball is thrown vertically upwards it was observed twice at a height of 'h' and a time interval of 't'.calculate initial velocity of body in terms of 'g' 't' and 'h'

ANS=1/2root(g^2*t^2-8gh) .......?

if correct then please aproove ill post the solution

√take the up direction as +ve. here s=h.

 substituting the value values in s=ut+1/2at²

h=ut-gt²

gt²-ut+h=0

t=u+√u²-4gh/2g    and   u-√u²-4gh/2g 

now the time is given t...

u+√u²-4gh/2g -   u-√u²-4gh/2g =t

solving this u²=g²t²+4gh

therefore u=√g²t²+4gh.