Math: a rocket is launched into the air from a machine 12m above the ground...?

^ the height, h(t)= -5t+30t+12, where t is time in seconds.

a.) what is the maximum height the rocket reaches?

b.) when will the rocket reach the ground?

I can’t factor it, it’s impossible so i need help solving this!! pls 

Update:

it would be really helpful if someone could maybe upload a picture of what they’re doing, you don’t have to at all, it’s fine if u just wanna comment the solution because anything helps :)

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  • 1 month ago
    Favourite answer

    h(t) = -5t² + 30t + 12

    or, h(t) = -5(t² - 6t) + 12

    Hence, h(t) = -5(t - 3)² + 57

    The maximum value of h(t) occurs when t = 3, i.e. h = 57

    The rocket hits the ground when h(t) = 0

    i.e. -5(t - 3)² + 57 = 0

    => (t - 3)² = 57/5

    so, t - 3 = √(57/5)

    or, t = 3 ± √(57/5)

    Hence, t = 6.4 seconds

    See the sketch below.

    :)>

    Attachment image
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    • Ash
      Lv 7
      1 month agoReport

      h(t) = -5(t² - 6t) + 12
      h(t) = -5(t² - 6t) -45 + 45 + 12
      h(t) =-5(t² - 6t + 9) + 57
      h(t) =-5(t -3)² + 57

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  • alex
    Lv 7
    1 month ago

    Hint:

    h(t) = -5t^2+30t+12 = -5(t-3)^2 + 57

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  • Ash
    Lv 7
    1 month ago

    You seem to have forgotten the exponent '2' on 't' of -5t

    h(t)= -5t²+30t+12

    velocity, v(t) = h'(t) = -10t + 30

    At max height v(t) = 0

    -10t + 30 = 0

    t = 3s  ↢ rocket reaches max height in 3 s

    max height = h(3) = -5(3)²+30(3)+12 = 57m

    When rocket reaches the ground h(t) = 0

    -5t²+30t+12 = 0

    5t²-30t-12 = 0

    Using quadratic solution

    t = {-(-30) ±√[(-30)²-4(5)(-12)]}/(2*5)

    t = (30±√1140)/10

    t = 6.4 s  ( ignore the other root which is negative)

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  • 1 month ago

    Edit: I have your Update.

    Trajectory is parabolic.

    Parabola use a standard form:

    y = ax^2 + bx + c 

    and solve for x with the "quadratic formula"

    Assuming -5t^2 + 30t + 12

    Vertex is -b/2a = -30/10 = 3 seconds 

    When t = 3 

    h = -45 + 90 + 12 = 57

    For flight time solve -5t^2 + 30t + 12 = 0

    time = 6.376 seconds 

    - 5x2 + 30x + +12 = 0

    Roots: 6.3763886032268, -0.37638860322683

    Root Pair: 3 ± 2√(285)/10

    Factored: f(x) = -5(x - 6.3763886032268)(x + 0.37638860322683)

    Discriminant: 1140

    Vertex: (3, 57)

    Sum of Roots (-b/a): 6

    Product of Roots

    Attachment image
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  • Is it -5t^2 + 30t + 12?

    -5t^2 + 30t + 12 =>

    -5 * (t^2 - 6t) + 12 =>

    -5 * (t^2 - 2 * 3 * t) + 12 =>

    -5 * (t^2 - 2 * 3 * t + 3^2 - 3^2) + 12 =>

    -5 * (t^2 - 6t + 9) + 5 * 3^2 + 12 =>

    -5 * (t - 3)^2 + 45 + 12 =>

    57 - 5 * (t - 3)^2

    57 - 5 * (t - 3)^2

    t = 3, we're at the critical point

    57 - 5 * (3 - 3)^2 =>

    57 - 5 * 0^2 =>

    57 - 5 * 0 =>

    57 - 0 =>

    57

    0 = 57 - 5 * (t - 3)^2

    5 * (t - 3)^2 = 57

    25 * (t - 3)^2 = 285

    5 * (t - 3) = +/- sqrt(285)

    t - 3 = +/- sqrt(285) / 5

    t = 3 +/- sqrt(285)/5

    t = (15 +/- sqrt(285)) / 5

    t = (15 + sqrt(285)) / 5

    • Timmy1 month agoReport

      on lines 3 and 4 that u wrote, im a little confused by the steps you’re taking, could u clarify please?

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