For firework launched from height 100ft with initial velocity 150ft/sec, equation made is correct
(a) equation will be 
(b) Now we have to see when it will land. At land or ground level height h will be equal to 0. So simply plug 0 in h place in equation made in part (a)

Now we have to solve this quadratic. We will use quadratic formula method to solve this equation.

a = -16, b = 150, c = 100.
Plugging these values in quadratic formula we get




time cannot be negative so we will drop this answer
then 
So 10 seconds is the answer for this
(c) To make table simply plug various value for t like t =0, 2, 4, 6, 8 till 10. Plug values in equation mad in part (a) and find h value for each t as shown
For t =0 seconds,
For t =2 seconds, 
For t =4 seconds, 
For t =6 seconds, 
For t =8 seconds,
For t =10 seconds, 
(d) Axis of symmetry is given by formula


t = 4.6875 is axis of symmetry line
(e) x-coordinate of vertex is again given by formula

so t = 4.6875
then to find y coordinate we will plug this value of t as 4.6875 in equation made in part (a)
For t =4.6875, 
so vertex is at (4.6875, 451.563)
(f) As the firework is launched so in starting time is t=0, we cannot have time before t=0 (negative values) practically. Also we cannnot have firework going down into the ground so we cannot have h value negative physically.