Answer: 1.106 s
Step-by-step explanation:
This situation is related to projectile motion and one the equation that models the height of the blueberry pie in time is:
![y=y_{o}+V_{o}sin(\theta) t-\frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%2BV_%7Bo%7Dsin%28%5Ctheta%29%20t-%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D)
Where:
is the blueberry pie final height (when it hits the ground)
is the blueberry pie initial height
is the blueberry pie initial velocity
is the angle, assuming the pie was shot horizontally
is the time
is the acceleration due gravity
Rewriting the equation:
![0=y_{o}-\frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=0%3Dy_%7Bo%7D-%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D)
Isolating
:
![t=\sqrt{\frac{2y_{o}}{g}}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2y_%7Bo%7D%7D%7Bg%7D%7D)
![t=\sqrt{\frac{2(6 m)}{9.8 m/s^{2}}}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2%286%20m%29%7D%7B9.8%20m%2Fs%5E%7B2%7D%7D%7D)
Finally:
![t=1.106 s](https://tex.z-dn.net/?f=t%3D1.106%20s)
First I find the individual widget weight
81 - 73 = 2x
8 = 2x
x = 4
Then I find how many widgets were left
73 - 1 = 4y
72 = 4y
18 = y
Y is the amount of widgets left, therefore the amount left is 18
12/3 = x/8
Cross multiply
3x = 96
Eliminate
3x/3 = 96/3
x = 32
Answer: A (-2, 1)
see image explanation attached