X=-5 a vertical line only has an X intercept
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.
Answer:
![312.5\pi \text{ km}^3\approx 981.75\text{ km}^3](https://tex.z-dn.net/?f=312.5%5Cpi%20%5Ctext%7B%20km%7D%5E3%5Capprox%20981.75%5Ctext%7B%20km%7D%5E3)
Step-by-step explanation:
We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.
Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.
Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.
, where,
r = Radius of cylinder,
h = height of the cylinder.
Upon substituting our given values in volume formula we will get,
![\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20a%20semicircular%20cylinder%7D%3D%5Cfrac%7B%5Cpi%20%285%5Ctext%7B%20km%7D%29%5E2%2A25%5Ctext%7B%20km%7D%7D%7B2%7D)
![\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20a%20semicircular%20cylinder%7D%3D%5Cfrac%7B%5Cpi%2A25%5Ctext%7B%20km%7D%5E2%2A25%5Ctext%7B%20km%7D%7D%7B2%7D)
![\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20a%20semicircular%20cylinder%7D%3D%5Cfrac%7B%5Cpi%2A625%5Ctext%7B%20km%7D%5E3%7D%7B2%7D)
![\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20a%20semicircular%20cylinder%7D%3D%5Cpi%2A312.5%5Ctext%7B%20km%7D%5E3)
![\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20a%20semicircular%20cylinder%7D%3D%5Cpi%2A312.5%5Ctext%7B%20km%7D%5E3)
![\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20a%20semicircular%20cylinder%7D%3D981.74770%5Ctext%7B%20km%7D%5E3)
Therefore, the volume of earth removed to build the three tunnels is
.