An object is propelled vertically upward from the top of a 144-foot building. The quadratic function s(t)= -16t2+192+144 models the ball's height above the ground, in feet, s(t) seconds after it was thrown. How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if necessary.
Answer
0.7 seconds
when it hits the ground, s(t) = 0
-16t^2 + 192t + 144 = 0
t^2 - 12t - 9 = 0
I will complete the square rather than use the quadratic formula
t^2 - 12t + 36 = 9 + 36
(t-6)^2 = 45
t-6 = ±√45
t = 6 ± √45
= appr -.7 or 12.7
since t > 0, it will hit the ground 12.7 after it was tossed.
Well, it must have really wanted to hit the ground quickly because it only took 0.7 seconds! I guess it was in a hurry to join the rest of its fellow objects down there.
To find the time it takes for the object to hit the ground, we need to determine when the height, given by the quadratic function, is equal to zero.
The quadratic function given is s(t) = -16t^2 + 192t + 144.
Setting s(t) = 0, we have:
-16t^2 + 192t + 144 = 0.
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = -16, b = 192, and c = 144.
t = (-192 ± √(192^2 - 4(-16)(144))) / (2(-16)).
Simplifying further:
t = (-192 ± √(36864 + 9216)) / (-32).
t = (-192 ± √46080) / (-32).
Now, we can calculate the solutions:
t = (-192 + √46080) / (-32) ≈ 0.7 seconds.
t = (-192 - √46080) / (-32) ≈ -4.5 seconds.
Since time cannot be negative in this context, we discard the negative solution.
Therefore, it takes approximately 0.7 seconds for the object to hit the ground.
To find the number of seconds it takes until the object hits the ground, we need to find when the height above the ground, given by the function s(t), is equal to 0.
The quadratic function s(t) = -16t^2 + 192t + 144 models the height above the ground in feet, where t is the time in seconds.
To find when the object hits the ground, we set s(t) equal to 0 and solve for t:
0 = -16t^2 + 192t + 144
Next, we can solve this quadratic equation using any appropriate method, such as factoring, completing the square, or using the quadratic formula.
However, in this case, the equation does not factor easily, and using the quadratic formula would be the most suitable approach.
The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -16, b = 192, and c = 144. Substituting these values into the quadratic formula, we get:
t = (-192 ± √(192^2 - 4(-16)(144))) / (2(-16))
This simplifies to:
t = (-192 ± √(36864 + 9216)) / (-32)
t = (-192 ± √(46080)) / (-32)
t = (-192 ± 214.926) / (-32)
We have two possible solutions:
t = (-192 + 214.926) / (-32) ≈ 0.71 seconds
t = (-192 - 214.926) / (-32) ≈ -6.04 seconds
The negative solution, t ≈ -6.04 seconds, does not make physical sense since time cannot be negative in this context. Therefore, we discard this solution.
Thus, the object finally hits the ground approximately 0.71 seconds after it was thrown.
Rounding to the nearest tenth of a second, the answer is 0.7 seconds.