Distance Calculus Question - Velocity and Derivatives, Parametric?

Question

please view second problem @ http://www.market2people.com/ec1.pdf

I cant figure this out. Thanks in advance for your help.

Answer 1

for 0 so its position at t=3 is
h= 144 -63=81m
v=33m/s
After t= 3s the movement is defined by
h= -5t^2+33t+81
h´= -10t+33 = 0 so t = 3.3s and hmax =135.45m