Mathematics, 03.11.2019 20:31 Josephjenkins620

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

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
problem.
g. on the same coordinate plane from #1, draw a graph that represents
the path of this firework. m

Answers: 1

Show answers

Answers

Answer from: darlaaa

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

h(t) = -16t^2 + 150t + 100

(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)

0 = -16t^2 + 150t +100

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

$\frac{0}{2} = \frac{-16t^2}{2} + \frac{150t}{2} + \frac{100}{2}$

0 = -8t^2 + 75t + 50

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

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ -----------------------------(1)

on comparing 0 = -8t^2 + 75t + 50 with 0 =ax^2 +bx +c we get

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

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

$t = \frac{-75\pm \sqrt{75^2-4(-8)(50)}}{2(-8)}$

$t = \frac{-75\pm \sqrt{7225}}{-16}$

$t = \frac{-75\pm \85}}{-16}$

$t = \frac{-75+ 85}}{-16}$ or $t = \frac{-75-85}}{-16}$

$t = \frac{10}}{-32}$ or $t = \frac{-160}}{-16}$

$t = \frac{-5}{8}$ or t = 10

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

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

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

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

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

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

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

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

so plug values of a and b from (2)

$t = \frac{-75}{2(-8)}$

$t = \frac{-75}{-16}$

t = 4.6875

So the axis of symmetry line equation is t = 4.6875

(e) Vertex formula is again given by

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

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 = -16(4.6875)^2+150(4.6875)+100

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