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Ann [662]
3 years ago
15

A. 22.1 ft^2 B. 20 ft^2 C. 13.5 ft^2 D. 16.8 ft^2

Mathematics
1 answer:
USPshnik [31]3 years ago
8 0
The answer would be D, 16.8 ft.^2
4 x 2.5 + 2^2 x 3.14 x .5= 16.8
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The required value of x is 4.

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Please help! ive been stuck on this for so long and i just keep getting frustrated can someone help walkme through this?
bagirrra123 [75]

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

(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

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}

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)}

t = \frac{-150 \pm  \sqrt{22500+6400}}{-32}

t = \frac{-150 \pm  \sqrt{28900}}{-32}

t = \frac{-150+170}{-32}  = \frac{20}{-32} = -0.625

time cannot be negative so we will drop this answer

then t = \frac{-150-170}{-32}  = \frac{-320}{-32} = 10

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, h = -16(0)^2+150(0)+100 = 100 feet

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

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

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

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

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

(d) Axis of symmetry is given by formula

x = \frac{-b}{2a}

t = \frac{-150}{2(-16)} =\frac{-150}{-32} = 4.6875

t = 4.6875 is axis of symmetry line

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

x = \frac{-b}{2a}

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

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.

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