Ive been stuck on this for so long and i just keep getting frustrated can someone

(a) Equation for firework launched upward with intial velocity 150 ft/sec from 100 ft high building is written correctly by you in comment.

(b) To find time when it will land the ground, we will plug 0 in h place as at ground level height h =0. So plug 0 in h place in equation made in part (a)

We can simplify this equation by dividing equation by greatest common factor in all terms which will be 2 here.

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

-----------------------------(1)

on comparing with we get

a = -8, b = 75, c = 50 ----------------------------------------------(2)

Plugging these values in quadratic formula equation (1) we get

or

or

or

We cannot have time as negative so answer will be t = 10 seconds

(C) To make table we will take few value of time like t=0,2,4,6,8,10 and plug them in equation made in part (a) to find h values

For t =0,

For t =2,

For t =4,

For t =6,

For t =8,

For t =10,

(d) Axis of symmetry of parabola is given by formula

so plug values of a and b from (2)

t = 4.6875

So the axis of symmetry line equation is t = 4.6875

(e) Vertex formula is again given by

so we get x coordinate as 4.6875

For y coordinate simply plug 4.6875 in t place in equation made in part (a)

h = 451.563

So vertex coordinate is (4.6875, 451.563)

(f) Firework is launched at time t=0, so we cannot go below 0, time has to be positive, we cannot have negative time physically. Also firework cannot go below the ground. so height cannot be negative also

(g) use all points of (c) to plot the sketch as shown in attachment.

So plot points (0,100); (2,336); (4,444); (6,424); (8,276); (10,0). Also vertex point at (4.6,451.5) approx