Calculating Kinetic Energy of a Ball: Is it Possible?

In summary: In this case, you must use the general equation for kinetic energy, KE = mv^2. Can you explain further? I don't really understand.The answer can be obtained by applying the conservation of energy. This equation states that the total energy of a system remains the same over time. Therefore, the total energy of the system must be equal to the final total energy. OK so I can do 2.3*9.8*15= 338.1 J which is final PE but since they're equal that is the answer for initial KE?Yes, that answer is correct. It is more correct to say that that is the "potential energy gained" rather than "final
  • #1
NatalieWise123
14
0

Homework Statement


A ball with a mass of 2.3 kg is thrown into the air. If the ball rises to a maximum height of 15 m before stopping and coming back down, how much kinetic energy did the ball have initially?

Homework Equations


KE=0.5*mass*velocity^2

The Attempt at a Solution


I don't think it's possible because no velocity is given and you can't find it because there is no time given. Am I missing something?[/B]
 
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  • #2
Why don't you just aply the conservation of the energy?

[tex]E_{i}=E_{f}[/tex]
 
  • #3
AhmirMalik said:
Why don't you just aply the conservation of the energy?

[tex]E_{i}=E_{f}[/tex]

Can you explain further? I don't really understand.
 
  • #4
The total energy of the system has to be conserved. That means, the total initial energy must be equal to the final total energy.

[tex]\frac{m(v_{i})^2}{2}+mgh_{i}=\frac{m(v_{f})^2}{2}+mgh_{f}[/tex]

where [tex]\frac{m(v_{i})^2}{2}[/tex] is what you are looking for.
 
  • #5
Exactly as above. You have to remember that energy is conserved, so the potential energy gained by the ball being thrown up, is equal to the kinetic energy lost and vice versa. Essentially in this problem, your answer can be obtained using [itex]KE_i = PE_f [/itex] Which is the same as what the person above did, but by taking the hand as the inial starting reference fram where PE=0, some terms fall out, leaving you with [itex]\frac{1}{2}mv_i^2 = mgh_f [/itex]
 
  • #6
FaraDazed said:
Exactly as above. You have to remember that energy is conserved, so the potential energy gained by the ball being thrown up, is equal to the kinetic energy lost and vice versa. Essentially in this problem, your answer can be obtained using [itex]KE_i = PE_f [/itex] Which is the same as what the person above did, but by taking the hand as the inial starting reference fram where PE=0, some terms fall out, leaving you with [itex]\frac{1}{2}mv_i^2 = mgh_f [/itex]

OK so I can do 2.3*9.8*15= 338.1 J which is final PE but since they're equal that is the answer for initial KE?
 
  • #7
NatalieWise123 said:
OK so I can do 2.3*9.8*15= 338.1 J which is final PE but since they're equal that is the answer for initial KE?
Yes that answer is correct. It is more correct to say that that is the "potential energy gained" rather than "final potential energy" as it is technically a change in energy, but that's just semantics I guess. But yes, since energy is conserved it is equal to the kinetic energy you are looking for.
 
  • #8
Can it be said that when components that contain a like kind of kinetic energy assemble together, the sum of that energy is shared between them? Abstractly it seems that this effect can be observed in every kind of cohesive bond within a physical transfer such as thermal transfer, pressure transfer and density transfers.
 
  • #9
NatalieWise123 said:

Homework Statement


A ball with a mass of 2.3 kg is thrown into the air. If the ball rises to a maximum height of 15 m before stopping and coming back down, how much kinetic energy did the ball have initially?

Homework Equations


KE=0.5*mass*velocity^2

The Attempt at a Solution


I don't think it's possible because no velocity is given and you can't find it because there is no time given. Am I missing something?[/B]
Yes, you are. The ball doesn't simply jump out of your hand. It must be propelled upward with an initial velocity. Knowing how far up the ball travels allows one to calculate the initial velocity. Sometimes, not all the facts are presented in neat packages suitable for plugging and chugging.
 

1. How do you calculate the kinetic energy of a ball?

To calculate the kinetic energy of a ball, you will need to know its mass (in kilograms) and velocity (in meters per second). The formula for kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. This formula can be used for any object in motion, including a ball.

2. Is it possible to calculate the kinetic energy of a ball in motion?

Yes, it is possible to calculate the kinetic energy of a ball in motion. As long as you know the mass and velocity of the ball, you can use the formula KE = 1/2 * m * v^2 to find its kinetic energy.

3. How is kinetic energy different from potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or condition. In the case of a ball, kinetic energy is the energy it has while it is moving, whereas potential energy would be the energy it has while it is at rest at the top of a hill or on a ledge.

4. Can the kinetic energy of a ball change?

Yes, the kinetic energy of a ball can change. It can change when the ball's mass or velocity changes. For example, if you throw a ball with a greater force, its velocity will increase, thus increasing its kinetic energy. Similarly, if you double the mass of the ball, its kinetic energy will also double.

5. Why is it important to calculate the kinetic energy of a ball?

Calculating the kinetic energy of a ball is important because it helps us understand the energy associated with an object in motion. This can be useful in various fields of science, such as physics and engineering, where understanding an object's energy can help in designing and predicting its behavior. Additionally, knowing the kinetic energy of a ball can also be important for safety reasons, such as in sports where the speed and impact of a ball can affect players' safety.

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