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
Well, let's solve this explosive problem together, shall we?
To find out how long it takes for the debris to hit the ground, we need to determine when the height, h(t), reaches zero. So, let's set the equation equal to zero:
-12(t+2)^2 + 120(t+2) = 0
Now, let's simplify and solve for t:
-12(t^2 + 4t + 4) + 120t + 240 = 0
-12t^2 - 48t - 48 + 120t + 240 = 0
-12t^2 + 72t + 192 = 0
Divide the equation by -12 to make it easier to work with:
t^2 - 6t - 16 = 0
Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = -6, and c = -16. Substituting those values, we get:
t = (-(-6) ± √((-6)^2 - 4(1)(-16))) / (2(1))
Simplifying further:
t = (6 ± √(36 + 64)) / 2
t = (6 ± √100) / 2
t = (6 ± 10) / 2
Now, we have two potential solutions:
t1 = (6 + 10) / 2 = 16 / 2 = 8
t2 = (6 - 10) / 2 = -4 / 2 = -2
Since time cannot be negative in this context, we disregard t2 and conclude that it will take approximately 8 seconds for the piece of debris to hit the ground after the explosion.
Remember, always approach explosive situations with caution and a healthy sense of humor!
To find out how long it will take for the debris to hit the ground, we need to find the value of t when h(t) = 0.
Given that h(t) = -12(t+2)^2 + 120(t+2), we can set h(t) = 0:
0 = -12(t+2)^2 + 120(t+2)
Now, let's solve this equation step-by-step:
1. Distribute -12 into (t+2)^2 and 120(t+2):
0 = -12t^2 - 48t - 48 + 120t + 240
2. Combine like terms:
0 = -12t^2 + 72t + 192
3. Set the equation equal to zero:
-12t^2 + 72t + 192 = 0
4. Divide the equation by -12 to simplify:
t^2 - 6t - 16 = 0
5. Now, we can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = 1, b = -6, and c = -16.
Substituting these values into the quadratic formula:
t = (-(-6) ± √((-6)^2 - 4(1)(-16))) / (2(1))
t = (6 ± √(36 + 64)) / 2
t = (6 ± √100) / 2
t = (6 ± 10) / 2
6. Simplify both solutions:
a) t = (6 + 10) / 2 = 16 / 2 = 8
b) t = (6 - 10) / 2 = -4 / 2 = -2
So, we have two values for t: t = 8 and t = -2.
However, since time cannot be negative in this context, we discard the negative value.
Therefore, it will take 8 seconds for the piece of debris to hit the ground after the explosion.
To find the time it takes for the piece of debris to hit the ground, we need to find the value of t when h(t) is equal to zero, since the height of the debris is zero at ground level.
Given the equation h(t) = -12(t+2)^2 + 120(t+2), we can set it equal to zero:
0 = -12(t+2)^2 + 120(t+2)
Now, let's simplify the equation:
0 = -12(t^2 + 4t + 4) + 120(t+2)
0 = -12t^2 - 48t - 48 + 120t + 240
0 = -12t^2 + 72t + 192
Next, we can divide the entire equation by -12 to simplify it further:
0 = t^2 - 6t - 16
Now, we have a quadratic equation. To solve for t, we can factorize or use the quadratic formula.
Since the quadratic equation does not easily factorize, we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = -6, and c = -16. Plugging these values into the quadratic formula:
t = (-(-6) ± √((-6)^2 - 4(1)(-16))) / (2(1))
t = (6 ± √(36 + 64)) / 2
t = (6 ± √100) / 2
t = (6 ± 10) / 2
We have two possible values for t:
1) t = (6 + 10) / 2
t = 16 / 2
t = 8
2) t = (6 - 10) / 2
t = -4 / 2
t = -2
Since time cannot be negative in this context, we discard the negative value.
Therefore, it will take 8 seconds for the piece of debris to hit the ground after the explosion.
it hits the ground when h=0. So, just solve
-12(t+2)^2 + 120(t+2) = 0
or, if you prefer
t^2-6t-16 = 0