Kinetic/potential energies for a ball thrown down

[Kork]

Hi again!

So, this time I have a ball with a mass m that I throw down from some point with a height h.

I want to write the kinetic and potentiel energies that the ball has in the start and in the end for this ball that is falling to h=0.

What I have thought of is that the ball must have a potential energy, u = mgh before it is thrown and when it hits the grown, and when it's moving down the energy is kinetic. I don't understand completely why and when it's undergoing a certain form of energy.

How does mgh=1/2mv^2 relate to this situation?

In my notes I also have that E = mgh + 1/2mv2 , which for some reason is the start energy?

 

[Kork]

Or is this correct:

In the start I have that:
E = mgh + 1/2mv^2

What about when it hits the ground?

 

[Doc Al]

Or is this correct:

In the start I have that:
E = mgh + 1/2mv^2

Yes, where h is the initial height above the ground and v the initial speed at which you throw the ball.

What about when it hits the ground?

Energy is conserved, so that potential energy gets transformed to additional kinetic energy.

The same equation applies, only now the height is the final height and the velocity is the final velocity. The total energy remains the same as it was at the start.

 

[emailanmol]

The basic concept is that mechanical energy of a body is constant if no external force does work on it.

In the start just before you throw the ball, the ball was at height h above the ground and had speed 0.
When you throw it you apply a force which does work on it and instantaneously makes the ball have speed v.

From here, till the ball reaches the ground the only external force acting on it is the gravity.

Now consider the ball and Earth as a system so that gravity is now an internal conservative force and regard the energy due to it as potential energy.
Thus no external force acts on this system (till it hits the ground)and its total mechanical energy i.e sum of potential and kinetic energies is constant.

Initially the ball was at height h above the ground(where we regard potential energy to be 0).
So its Total E was mv^2/2 + mgh.

Later when it reached the ground (just before touching) suppose its speed becomes x.here height is 0 and body has only kinetic energy . Thus mx^2/2 = E.

 

[PJC01]

I can definitely help explain the concept of kinetic and potential energy in this scenario. First, let's define these two types of energy.

Kinetic energy is the energy an object possesses due to its motion. It is given by the equation: KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

Potential energy, on the other hand, is the energy an object possesses due to its position or state. In this case, the potential energy is gravitational potential energy, which is given by the equation: PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

Now, let's apply these concepts to the ball being thrown down. At the start, when the ball is held at a height h, it has potential energy due to its position. This potential energy is given by mgh. As the ball is thrown, it begins to move and gains kinetic energy. This kinetic energy is given by 1/2mv^2. As the ball falls and reaches the ground, its potential energy decreases to 0, while its kinetic energy reaches its maximum value. This is because all of the potential energy has been converted into kinetic energy as the ball falls. This is represented by the equation: mgh = 1/2mv^2.

In your notes, you have E = mgh + 1/2mv^2, which represents the total energy of the ball at any given point during its fall. This is because the ball has both potential and kinetic energy simultaneously throughout its fall. As the ball falls, its potential energy decreases while its kinetic energy increases, but the total energy (E) remains constant.

I hope this helps to clarify the relationship between mgh and 1/2mv^2 in this situation. It is important to note that these equations are specific to objects moving in a gravitational field and may not apply to other forms of energy conversion. Keep exploring and asking questions to deepen your understanding of these concepts!