Let T and H be the total time of flight and height respectively.

Then H = 0.5 * g * T^2

Let v m/s be the velocity of the particle at the beginning of last  2 second.

v = g*(T-2)

according  to the problem, 176.4 = v*2 + 0.5 * g  * 2^2

                                           = 4gT -2g

                               => T = (176.4 + 2*9.8)/(4*9.8)

                                       = 5

Hence H = 0.5 * 9.8 * 25 = 122.5

let u be the vel. before the 2 sec. then.........

         s=176.4    a=9.9         t=2s

         s=ut+1/2at2

         176.4=u*2+1/2*9.8*4

         u=78.4 m/s

this u will be the final velocity for the block when dropped from heighest point before last 2 sec....

 then,

       u=0  ,v=78.4   ,a=9.8,  s=?

       2as=v*2-u*2

       on calculating   s=313.6m

       total h=313.6+176.4=490m