Answer:
Velocity=14[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.
In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.
![Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3D3%5Bkg%5D%5C%5Ch%3D20%5Bm%5D%5C%5C)
![Ep=3*9.81*20\\Ep=588.6[J]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A20%5C%5CEp%3D588.6%5BJ%5D)
When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.
![Ep=3*9.81*10\\Ep=294.3[m]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A10%5C%5CEp%3D294.3%5Bm%5D)
If we know that potential energy is transformed into kinetic energy, we can find the value of speed.
![Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Ctherefore%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%2A2%7D%7Bm%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B294.3%2A2%7D%7B3%7D%20%7D%20%5C%5C%5C%5Cv%3D14%5Bm%2Fs%5D)
<span>particle varies with time as shown in the diagram. ... resultant has a magnitude equal to 8.0. .... A constant force F is applied to a body of mass m that initially is headed east at velocity .... If the resultant force acting on a 2.0-kg object is equal to ..... A ball of mass mB is released from rest and acquires velocity of magnitude vB ...</span><span>
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