Answer:
![v_{o}=8.85m/s](https://tex.z-dn.net/?f=v_%7Bo%7D%3D8.85m%2Fs)
Explanation:
To determine the <u>muzzle velocity of the gun</u>, we must know how long does it take the ball to <u><em>strikes the ground </em></u>
![y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%2Bv_%7Boy%7Dt%2B%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D)
Since the ground is at y=0 and ![v_{oy}=0](https://tex.z-dn.net/?f=v_%7Boy%7D%3D0)
![0=1-\frac{1}{2}(9.8)t^{2}](https://tex.z-dn.net/?f=0%3D1-%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E%7B2%7D)
Solving for t
![t=0.4517s](https://tex.z-dn.net/?f=t%3D0.4517s)
Now, to determine the muzzle velocity we need to find its acceleration first
(1)
(2)
If we analyze the final velocity is 0. From (2) we have that
(3)
Replacing (3) in (1)
![2=-at^{2}+\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=2%3D-at%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
![2=a(0.4517)^{2} (\frac{1}{2}-1)](https://tex.z-dn.net/?f=2%3Da%280.4517%29%5E%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%7D-1%29)
![a=-19.60m/s^{2}](https://tex.z-dn.net/?f=a%3D-19.60m%2Fs%5E%7B2%7D)
Solving (3)
![v_{ox}=-at=-(19.60m/s^{2} )(0.4517s)=8.85m/s](https://tex.z-dn.net/?f=v_%7Box%7D%3D-at%3D-%2819.60m%2Fs%5E%7B2%7D%20%29%280.4517s%29%3D8.85m%2Fs)
Answer:
So kinetic means to move, something like that right, so the two balls that go in the air are where the kinetic energy is.
Explanation:
Hope it helps.
Answer:
0.22m/s
Explanation:
The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.
momentum of an object = mass* velocity
Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;
Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;
Solve for x.