For firework launched from height 100ft with initial velocity 150ft/sec, equation made is correct
(a) equation will be ![h(t) = -16t^2+150t+100](https://tex.z-dn.net/?f=%20h%28t%29%20%3D%20-16t%5E2%2B150t%2B100%20)
(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)
![0 = -16t^2 + 150t + 100](https://tex.z-dn.net/?f=%200%20%3D%20-16t%5E2%20%2B%20150t%20%2B%20100%20)
Now we have to solve this quadratic. We will use quadratic formula method to solve this equation.
![t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%20%20)
a = -16, b = 150, c = 100.
Plugging these values in quadratic formula we get
![t = \frac{-150 \pm \sqrt{150^2-4(-16)(100)}}{2(-16)}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-150%20%5Cpm%20%20%5Csqrt%7B150%5E2-4%28-16%29%28100%29%7D%7D%7B2%28-16%29%7D%20%20%20)
![t = \frac{-150 \pm \sqrt{22500+6400}}{-32}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-150%20%5Cpm%20%20%5Csqrt%7B22500%2B6400%7D%7D%7B-32%7D%20%20%20)
![t = \frac{-150 \pm \sqrt{28900}}{-32}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-150%20%5Cpm%20%20%5Csqrt%7B28900%7D%7D%7B-32%7D%20%20)
![t = \frac{-150+170}{-32} = \frac{20}{-32} = -0.625](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-150%2B170%7D%7B-32%7D%20%20%3D%20%5Cfrac%7B20%7D%7B-32%7D%20%3D%20-0.625%20)
time cannot be negative so we will drop this answer
then ![t = \frac{-150-170}{-32} = \frac{-320}{-32} = 10](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-150-170%7D%7B-32%7D%20%20%3D%20%5Cfrac%7B-320%7D%7B-32%7D%20%3D%2010%20%20)
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, ![h = -16(2)^2+150(2)+100 =336 feet](https://tex.z-dn.net/?f=%20h%20%3D%20-16%282%29%5E2%2B150%282%29%2B100%20%3D336%20feet%20)
For t =4 seconds, ![h = -16(4)^2+150(4)+100 = 444 feet](https://tex.z-dn.net/?f=%20h%20%3D%20-16%284%29%5E2%2B150%284%29%2B100%20%3D%20444%20feet%20)
For t =6 seconds, ![h = -16(6)^2+150(6)+100 = 424 feet](https://tex.z-dn.net/?f=%20h%20%3D%20-16%286%29%5E2%2B150%286%29%2B100%20%3D%20424%20feet%20)
For t =8 seconds,![h = -16(8)^2+150(8)+100 = 276 feet](https://tex.z-dn.net/?f=%20h%20%3D%20-16%288%29%5E2%2B150%288%29%2B100%20%3D%20276%20feet%20)
For t =10 seconds, ![h = -16(10)^2+150(10)+100 = 0 feet](https://tex.z-dn.net/?f=%20h%20%3D%20-16%2810%29%5E2%2B150%2810%29%2B100%20%3D%200%20feet%20)
(d) Axis of symmetry is given by formula
![x = \frac{-b}{2a}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-b%7D%7B2a%7D%20%20)
![t = \frac{-150}{2(-16)} =\frac{-150}{-32} = 4.6875](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B-150%7D%7B2%28-16%29%7D%20%3D%5Cfrac%7B-150%7D%7B-32%7D%20%3D%204.6875%20)
t = 4.6875 is axis of symmetry line
(e) x-coordinate of vertex is again given by formula
![x = \frac{-b}{2a}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-b%7D%7B2a%7D%20%20)
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, ![h = -16(4.6875)^2+150(4.6875)+100 = 451.563](https://tex.z-dn.net/?f=%20h%20%3D%20-16%284.6875%29%5E2%2B150%284.6875%29%2B100%20%3D%20451.563%20%20)
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.