1. Jan 29, 2007

### FlashMXHelp

At school we have a problem to solve but i dont know how to solve this problem.
Here is the condition:"Ball is throw vertical upward with inceptive speed v0.Of what height h the kinetic energy of the ball decrease two times? Resistance of the air is ignored." Im sorry if dont write it right but i dont know english very good.I need help please if anyone can help write the solution of the problem.

2. Jan 29, 2007

### cristo

Staff Emeritus
Is the question something like: "At what height is the kinetic energy equal to half of the initial kinetic energy?"

3. Jan 29, 2007

### FlashMXHelp

Nope its not like that.The answer of this problem is "h=v0(the null is index)2(2 is power)/4g
but i dont know how to get to the answer.

4. Jan 29, 2007

### cristo

Staff Emeritus
With that answer, it seems like the question is what I posted.

Ok, a few hints/questions:
1. What is the kinetic energy at the original position?
2. What does the total energy add up to at all points on the ball's trajectory?
3. What other energy does the ball have at the height you require?

5. Jan 31, 2007

### Virtual R

First find how the speed has changed when the Kinetic energy has decreased 2 times
So u see its 1/sqrt(2)times the original vo
then using the eq of motion 2gh=vf(2 in power) - vi (2 in power)
and putting vf=0 u can find h

6. Jan 31, 2007

### P3X-018

The expression for the height $h$ above the ground is actually a displacement and is sign sensitive. If you calculate positiv upwards, then the acceleration must be negative, since it is in the opposite direction.

As for the FlashMXHelp, a little different approach is as follows; the equation you need is as mentioned by Virtual,

$$-2gh = v^2 - v_0^2 \qquad\textrm{or}\qquad 2gh=v_0^2-v^2$$

Where $v$ is the velocity (and speed) at a height $h$ above the ground where the ball is fired. In addition to cristo's hints, here is another one,
What is the velocity (speed) of the ball at a given height $h$, using the above equation?
Compare this to the initial kinetic energy (Cristo's 1. hint).

7. Jan 31, 2007

### cristo

Staff Emeritus
I originally attempted to hint to the original poster how to solve the problem, since his post didn't include any work (the other people posting in this thread should probably read the guidelines re giving help when no work is shown!)
It is far more intuitive to consider the total energy at all points during the particle's flight. Thus $$\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv^2$$ Now, you know the relationship between 1/2mv02 and 1/2mv2, and so you can solve this for h.

Last edited: Jan 31, 2007