Answer:
Answer: LCM of 20,35 and 50 is 1400
Step-by-step explanation:
Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
![0=-16 t^{2} +122t+99](https://tex.z-dn.net/?f=0%3D-16%20t%5E%7B2%7D%20%2B122t%2B99)
Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
![-99=-16 t^{2} +122t+99](https://tex.z-dn.net/?f=-99%3D-16%20t%5E%7B2%7D%20%2B122t%2B99)
![-16 t^{2} +122+198=0](https://tex.z-dn.net/?f=-16%20t%5E%7B2%7D%20%2B122%2B198%3D0)
Now we can apply the quadratic formula
![t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-b%2Bor-%20%5Csqrt%7B%20b%5E%7B2%7D%20-4ac%7D%20%7D%7B2a%7D%20)
where a=-16, b=122, and c=198
![t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-122%2Bor-%20%5Csqrt%7B%20122%5E%7B2%7D-%284%29%28-16%29%28198%29%20%7D%20%7D%7B%282%29%28-16%29%7D%20)
![t= \frac{-122+ \sqrt{27556} }{-32}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-122%2B%20%5Csqrt%7B27556%7D%20%7D%7B-32%7D%20)
or
![t= \frac{-122- \sqrt{27556} }{-32}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-122-%20%5Csqrt%7B27556%7D%20%7D%7B-32%7D%20)
![t= \frac{-122+166}{-32}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-122%2B166%7D%7B-32%7D%20)
or
![t= \frac{-122-166}{-32}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-122-166%7D%7B-32%7D%20)
![t= \frac{-11}{8}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B-11%7D%7B8%7D%20)
or
![t=9](https://tex.z-dn.net/?f=t%3D9)
Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.