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A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at the edge of the building is v0, as shown. m h v0 m what is the kinetic energy of the rock just before it hits the ground? 1. kf = 1 2 m v2 0 2. kf = 1 2 m v2 0 − m g h 3. kf = 1 2 m v2 0 + m g h 4. kf = m g h − 1 2 m v2 0 5. kf = m g h

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skyluke89
The correct answer is 3) [tex]K_f = \frac{1}{2}mv_0^2 + mgh [/tex].

In fact, the total energy of the rock when it leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
[tex]E=U_i+K_i=mgh + \frac{1}{2}mv_0^2 [/tex]
As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy: 
[tex]E=K_f[/tex]
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
[tex]K_f = mgh + \frac{1}{2}mv_0^2 [/tex]
  • The kinetic energy of the rock just before hitting the ground is:

[tex]K_f=mgh+\frac{1}{2}mv_o^2[/tex]

  • Given information:

A rock of mass m is thrown horizontally off a building from a height h

As, the total energy of rock at the time of leaving the thrower's hand is the sum of gravitational potential energy and the initial kinetic energy.

[tex]E=U_i+K_i\\E=mgh+(1/2)mv_0^2[/tex]

As, the height from the ground decreases, the potential energy before hitting the ground is zero and the total final energy is just kinetic energy:

[tex]E=K_f[/tex]

As, the law of conservation of energy states the total final energy must be equal to the initial energy.

Hence, the final kinetic energy will be

[tex]K_f=mgh+\frac{1}{2}mv_o^2[/tex]

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https://brainly.com/question/17858145?referrer=searchResults