# 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

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 :)

### 5 Answers

- Wayne DeguManLv 71 month agoFavourite 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.

:)>

- AshLv 71 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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- Born YesterdayLv 71 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

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- 1 month ago
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

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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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