Calculating Kinetic Energy of a Ball: Is it Possible?

AI Thread Summary
To calculate the initial kinetic energy of a 2.3 kg ball thrown to a height of 15 m, the conservation of energy principle is applied. The potential energy at the maximum height, calculated as 338.1 J using the formula PE = mgh, is equal to the initial kinetic energy since energy is conserved. The discussion emphasizes that while velocity is not directly provided, it can be inferred from the height reached. Understanding that the ball must have an initial velocity to reach that height is crucial. Thus, the initial kinetic energy is determined to be 338.1 J.
NatalieWise123
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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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Why don't you just aply the conservation of the energy?

E_{i}=E_{f}
 
AhmirMalik said:
Why don't you just aply the conservation of the energy?

E_{i}=E_{f}

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

\frac{m(v_{i})^2}{2}+mgh_{i}=\frac{m(v_{f})^2}{2}+mgh_{f}

where \frac{m(v_{i})^2}{2} is what you are looking for.
 
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 KE_i = PE_f 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 \frac{1}{2}mv_i^2 = mgh_f
 
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 KE_i = PE_f 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 \frac{1}{2}mv_i^2 = mgh_f

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?
 
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.
 
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.
 
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.
 
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