Answer:
Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]
Explanation:
First, we have to break down the velocity vector into the X & y components.
![(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7B0%7D%20%3D%2015%20%2A%20cos%28%2035%29%3D%2012.28%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7B0%7D%20%3D%2015%20%2A%20sin%28%2035%29%3D%208.6%5Bm%2Fs%5D%5C%5C%5C%5C)
To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.
![y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]](https://tex.z-dn.net/?f=y%3D%28v_%7By%7D%20%29_%7B0%7D-%5Cfrac%7B1%7D%7B2%7D%2Ag%2At%5E%7B2%7D%20%20%20%5C%5Cwhere%3A%5C%5Cg%3D9.81%5Bm%2Fs%5E2%5D%5C%5Ct%20%3D%20time%5Bs%5D%5C%5Cy%3D0%5Bm%5D)
![0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]](https://tex.z-dn.net/?f=0%3D8.6%2At-%5Cfrac%7B1%7D%7B2%7D%2A9.81%2At%5E%7B2%7D%20%20%5C%5C4.905%2At%5E%7B2%7D%3D8.6%2At%5C%5C%20t%3D1.75%5Bs%5D)
In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.
![x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]](https://tex.z-dn.net/?f=x%20%3D%20%28v_%7Bx%7D%29_%7B0%7D%20%20%2At%5C%5Cx%3D12.28%2A1.75%5C%5Cx%3D21.5%20%5Bm%5D)
In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation
![y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5Cy%20%3D%200%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.81%2A%281.75%29%5E%7B2%7D%5C%5C%20y%3D15%20%5Bm%5D)
This question may require more information for an accurate answer. If the clouds are observed west of the city and they are moving in a westerly direction it means that the clouds are moving away. Precipitation refers to rainfall, sleet, hail or snow. Since no reference was made to the temperature, season or location of the city it could be rain or snow clouds. The clouds are moving away from the city so the possible answers could be b. or d. Furthermore, by dividing the distance by the speed one would get an answer of twenty four hours. So even if the clouds were moving towards the city it would take a day to reach and precipitation would not be likely in 24 hrs.
Isn't it "gravity" this would makes sense because grvaity difines weight
Answer:
The answer to your question is: C. -9.81 m/s²
Explanation:
A. 9.81 m/s² acceleration is considered positive when it goes to the center of the earth, so this option is incorrect.
B. 0 m/s² This option is incorrect because acceleration is 0 for a linear motion without acceleration.
C. -9.81 m/s² If a projectile goes to the sky, then the acceleration will be negative.
D. It is not constant. Acceleration is constant.
It’s coming in contact with more air molecules than I would if it was in a ball because there is less surface area
