An explosives company specializes in bringing down old buildings. The time it takes for a particular piece of flying debris to fall to the ground can be modelled by the function h(t)= -12(t+2)^2 + 120(t+2), where h is the height of the debris in metres and t is the time in seconds.

1.) How long will it take for the piece of debris to hit the ground after an explosion? Show all work

is there something unclear about the solution at

http://www.jiskha.com/display.cgi?id=1464662427

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To find the time it takes for the piece of debris to hit the ground, we need to determine when the height of the debris, represented by the function h(t), reaches 0.

The given function is h(t) = -12(t+2)^2 + 120(t+2).

Setting h(t) to zero and solving for t:

0 = -12(t+2)^2 + 120(t+2)

Next, let's simplify the equation:

0 = -12(t^2 + 4t + 4) + 120(t+2)
0 = -12t^2 - 48t - 48 + 120t + 240

Now, combine like terms:

0 = -12t^2 + 72t + 192

Now, we have a quadratic equation. Let's solve it by factoring or using the quadratic formula.

The equation is in the form: at^2 + bt + c = 0, where a = -12, b = 72, and c = 192.

Factoring seems a bit complex in this case, so let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values, we get:

t = (-(72) ± √((72)^2 - 4(-12)(192))) / (2(-12))
t = (-72 ± √(5184 + 9216)) / (-24)
t = (-72 ± √(14400)) / (-24)
t = (-72 ± 120) / (-24)

Simplifying further:

t = (-72 + 120) / (-24) or t = (-72 - 120) / (-24)
t = 48 / (-24) or t = -192 / (-24)
t = -2 or t = 8

From the solutions obtained for t, we discard the negative value since time cannot be negative in this context.

Therefore, it will take 8 seconds for the piece of debris to hit the ground after the explosion.