Please help! ive been stuck on this for so long and i just keep getting frustrated can someone help walkme through this?

i did letter a which i believe is h(t)=-16T^2+150 o T+100 o
and the height of the building is 100 feet.

One of the fireworks is launched from the top of the building with an initial
upward velocity of 150 ft/sec.
a. What is the equation for this situation?
b. When will the firework land if it does not explode?
c. Make a table for this situation so that it shows the height from time
t = 0 until it hits the ground.
d. Calculate the axis of symmetry.
e. Calculate the coordinates of the vertex.
f. Explain why negative values for t and h t( ) do not make sense for this

(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)

[tex] 0 = -16t^2 + 150t + 100 [/tex]

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

[tex] t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} [/tex]

a = -16, b = 150, c = 100.

Plugging these values in quadratic formula we get

[tex] t = \frac{-150 \pm \sqrt{150^2-4(-16)(100)}}{2(-16)} [/tex]

[tex] t = \frac{-150 \pm \sqrt{22500+6400}}{-32} [/tex]

[tex] t = \frac{-150 \pm \sqrt{28900}}{-32} [/tex]

[tex] t = \frac{-150+170}{-32} = \frac{20}{-32} = -0.625 [/tex]

time cannot be negative so we will drop this answer

then [tex] t = \frac{-150-170}{-32} = \frac{-320}{-32} = 10 [/tex]

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, [tex] h = -16(0)^2+150(0)+100 = 100 feet [/tex]

For t =2 seconds, [tex] h = -16(2)^2+150(2)+100 =336 feet [/tex]

For t =4 seconds, [tex] h = -16(4)^2+150(4)+100 = 444 feet [/tex]

For t =6 seconds, [tex] h = -16(6)^2+150(6)+100 = 424 feet [/tex]

For t =8 seconds,[tex] h = -16(8)^2+150(8)+100 = 276 feet [/tex]

For t =10 seconds, [tex] h = -16(10)^2+150(10)+100 = 0 feet [/tex]

(d) Axis of symmetry is given by formula

[tex] x = \frac{-b}{2a} [/tex]

[tex] t = \frac{-150}{2(-16)} =\frac{-150}{-32} = 4.6875 [/tex]

t = 4.6875 is axis of symmetry line

(e) x-coordinate of vertex is again given by formula

[tex] x = \frac{-b}{2a} [/tex]

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, [tex] h = -16(4.6875)^2+150(4.6875)+100 = 451.563 [/tex]

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.