That's going to depend on what the ordered pair of the original point Q
was ... an important piece of information that you forgot to give us.
Answer:
![t \approx 20.7\,s](https://tex.z-dn.net/?f=t%20%5Capprox%2020.7%5C%2Cs)
Step-by-step explanation:
Given that rock experiments a constant acceleration, position function can be obtained by integrating twice:
![v(t) = v_{o} - a\cdot t](https://tex.z-dn.net/?f=v%28t%29%20%3D%20v_%7Bo%7D%20-%20a%5Ccdot%20t)
![s(t) = s_{o} + v_{o}\cdot t -\frac{a}{2}\cdot t^{2}](https://tex.z-dn.net/?f=s%28t%29%20%3D%20s_%7Bo%7D%20%2B%20v_%7Bo%7D%5Ccdot%20t%20-%5Cfrac%7Ba%7D%7B2%7D%5Ccdot%20t%5E%7B2%7D)
The initial conditions of the rock are, respectively:
![s_{o} = 2100\,m](https://tex.z-dn.net/?f=s_%7Bo%7D%20%3D%202100%5C%2Cm)
![v_{o} = 0\,\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D%200%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Position of the rock as a function of time is:
![s(t) = 2100\,m -(4.9\,\frac{m}{s^{2}})\cdot t^{2}](https://tex.z-dn.net/?f=s%28t%29%20%3D%202100%5C%2Cm%20-%284.9%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20t%5E%7B2%7D)
The time taken for the rock to hit the canyon floor is:
![0\,m = 2100\,m - (4.9\,\frac{m}{s^{2}} )\cdot t^{2}](https://tex.z-dn.net/?f=0%5C%2Cm%20%3D%202100%5C%2Cm%20-%20%284.9%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20t%5E%7B2%7D)
![2100\,m = (4.9\,\frac{m}{s^{2}} )\cdot t^{2}](https://tex.z-dn.net/?f=2100%5C%2Cm%20%3D%20%284.9%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20t%5E%7B2%7D)
![t = \sqrt{\frac{2100\,m}{4.9\,\frac{m}{s^{2}} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cfrac%7B2100%5C%2Cm%7D%7B4.9%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%7D%20%7D)
![t \approx 20.7\,s](https://tex.z-dn.net/?f=t%20%5Capprox%2020.7%5C%2Cs)
Answer:
the answer would be c
Step-by-step explanation:
W=6 hopefully I helped!:)