A particle is projected vertically upwards from the ground so that its velocity (in m/s) is given by v1=128-32t (t lies between and including 0 to 4), and v2=3t^2-32t+80 where t>4, where t is the time in seconds after projection. Calculate

(i) the height of the particle when t=4,

(ii) the height of the particle when t=6,

(iii) the value of t when the acceleration of the particle is 40 m per sec per sec.

I can solve only part (i).

Answers given: (i) 256 m, (ii) 248 m and (iii) 12