Projectile motion and potential/kinetic energy

[golgiapparatus88]

I was doing a passage on projectile motion and this question threw me for a loop. Three balls are dropped from a cliff. One horizontal, one at an angle 30 above the horizontal and one 60 degrees above. The question asks which will have the greatest kinetic energy when they hit the ground? The answer says they all land with the same KE. that was my intuition but when I thought about it, the 60 degree ball reaches the highest point giving it a higher potential energy. since they each reach a different max height and all land at a lower position, shouldn't the 60 degree angle give a higher final KE?

I also just made up am example with two different angles and got two different KEs

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[attixx]

I don't know if I've worked that same question, but TBR has a problem that's similar. It talks about 3 cannons shooting a ball with a speed = v. One horizontal, one at 30 degrees and one at 60 degrees. This would be the same situation as in your problem, as long as it is known that all three balls are launched with the same speed

Conservation of energy states Ei=Ef, or KE + PE (initial) = KE + PE (final).

They all have the same speed. It doesn't matter if its all vertical, all horizontal, or partially horizontal and partially vertical (as in the case of launching at an angle). The also have the same initial potential energy, because they start at the same spot on the cliff at the same height. They hit the ground with all kinetic, and KEf = KEi + PEi. So if KEi is the same in all scenarios, and PEi is the same, then final KE has to be the same in all scenarios.

Energy is a scalar quantity, and will ultimately be conserved independent of the path taken (as long as nothing is lost to work).

 

[golgiapparatus88]

I don't know if I've worked that same question, but TBR has a problem that's similar. It talks about 3 cannons shooting a ball with a speed = v. One horizontal, one at 30 degrees and one at 60 degrees. This would be the same situation as in your problem, as long as it is known that all three balls are launched with the same speed

Conservation of energy states Ei=Ef, or KE + PE (initial) = KE + PE (final).

They all have the same speed. It doesn't matter if its all vertical, all horizontal, or partially horizontal and partially vertical (as in the case of launching at an angle). The also have the same initial potential energy, because they start at the same spot on the cliff at the same height. They hit the ground with all kinetic, and KEf = KEi + PEi. So if KEi is the same in all scenarios, and PEi is the same, then final KE has to be the same in all scenarios.

Energy is a scalar quantity, and will ultimately be conserved independent of the path taken (as long as nothing is lost to work).

Yup same problem! Thanks for the help

 

[MedPR]

Just did this passage problem. Why is KE=1/2mv^2 irrelevent? My thought process was KE=1/2mv^2, then use a=v/t and get KE=1/2m(ta)^2. Thus, whichever ball was in the air the longest would have the greatest kinetic energy.

The only explanation I can think of for that thought process being incorrect is that KE=1/2mv^2 is relevant only for instantaneous velocity? Is that even true?

Thanks.

 

[Ssina]

I wouldn't look at it that way... the important thing is that they all have the same initial velocity. Now the ball that was launched at an angle (has the same KE as the other balls at first).... goes up, so it is gaining PE and losing KE... until reaches the top, which all the KE from the initial velocity is converted to PE... then it will fall back down, so now we are doing the reverse, turning PE into KE.... and without any air resistance, by the time the ball gets to the original launch point, it has lost all of the extra PE that it gained and converted it into KE... once again we have the same velocity as when we started... so you didn't gain anything and didn't lose anything, and now it will fall from the same hight like the ball that was thrown straight down... so they all have the same KE at the point of impact!

working from OP intuition... the ball that reaches a higher height.... now has more PE (because it's has more h for mgh) but has no KE (since at top v=0m/s) vs. the ball that started from the original point that has mgh + KE... so essentially you lose that extra PE to gain it in the KE (as velocity) on the way down

 

[Pisiform]

The main idea as the above poster said is the conservation of energy.

Total Energy at the start = KE + PE = 0.5mv^2 + mgh ...(note that v is the resultant velocity and not the horizontal or vertical component of velocity.)

Total Energy at the end = PE + KE = mgh + 0.5mv^2 (v is resultant again)

so main idea is that KE depends on the resultant velocity and not their components. As long as resultant velocity is same, and all the balls are falling to the same level, then KE in the end is the same for all of them.

However,if they were asking you the KE at the top then that would be diff. for each object because at top point, Vertical velocity is 0. KE at the top would have been 0.5m(vcostheta)^2.

 

[Erythropoietin]

Lets say you are at a cliff that is 100 m high and the mass is 1 kg, so the PE will be mgh = 1000 J.
If you simply drop that object down, 1000 J of PE will transfer to 1000 J of KE when it hits the ground.

When you are at a 100 m height and you project the ball upward by some angle, you have to treat the height as the "ground level." Thus, you start with 1000 J KE, which is transferred to 1000 J PE at the max height of the projectile, which is then again transferred back to 1000 J KE when it is back on the "ground level" (i. e., at 100 m height). So what happened? Nothing. You started with 1000 J of energy when you projected it upward, and you ended with 1000 J when the ball reached the 100 m level again. So at that 100 m point (when the ball is going downwards), you still have 1000 J, which is no different than the amount of energy that was required to drop the object without projecting it.
Therefore, they all have the same potential energy, and thus the same kinetic energy at the end.

 

[MedPR]

Makes sense. Thanks for all the (great) responses.

 

[Pisiform]

Lets say you are at a cliff that is 100 m high and the mass is 1 kg, so the PE will be mgh = 1000 J.
If you simply drop that object down, 1000 J of PE will transfer to 1000 J of KE when it hits the ground.

When you are at a 100 m height and you project the ball upward by some angle, you have to treat the height as the "ground level." Thus, you start with 1000 J KE, which is transferred to 1000 J PE at the max height of the projectile, which is then again transferred back to 1000 J KE when it is back on the "ground level" (i. e., at 100 m height). So what happened? Nothing. You started with 1000 J of energy when you projected it upward, and you ended with 1000 J when the ball reached the 100 m level again. So at that 100 m point (when the ball is going downwards), you still have 1000 J, which is no different than the amount of energy that was required to drop the object without projecting it.
Therefore, they all have the same potential energy, and thus the same kinetic energy at the end.

I don not agree with this. If you are starting at a height of 100m and you project the ball upward by lets say 30 degrees with a velocity of 1m/s ... then the TE at that starting point will be KE + PE. You cannot ignore PE because in the end ball will be falling below the level from where u started.

So the TE in the start would be = 0.5mv^2 + mgh.

Now at the top part, you do not say that all the energy is PE because we still have horizontal component of velocity and hence some KE = 0.5m(vcosthta)^2. So PE at that point would be TE - KE_Horizontal

In the end, the Total energy would be equal to KE because the ball landed at ground level.

 

[Erythropoietin]

I don not agree with this. If you are starting at a height of 100m and you project the ball upward by lets say 30 degrees with a velocity of 1m/s ... then the TE at that starting point will be KE + PE. You cannot ignore PE because in the end ball will be falling below the level from where u started.

So the TE in the start would be = 0.5mv^2 + mgh.

Now at the top part, you do not say that all the energy is PE because we still have horizontal component of velocity and hence some KE = 0.5m(vcosthta)^2. So PE at that point would be TE - KE_Horizontal

In the end, the Total energy would be equal to KE because the ball landed at ground level.

When the projectile falls back to the "100 m" level , it's all 1000 J of energy anyway. So you will still have 1000 J of energy even if you projeced it at 1 degree, 45 degree or 80 degree

 

[Erythropoietin]

http://physics.bu.edu/~redner/211-sp06/class-energy/threeballs.html

 

[Pisiform]

When the projectile falls back to the "100 m" level , it's all 1000 J of energy anyway. So you will still have 1000 J of energy even if you projeced it at 1 degree, 45 degree or 80 degree

I know and I mentioned that. I am saying that 100 m would not be the ground level .... At 100 m ball would still be in the air because u started from a height.

 

[MedPR]

I know and I mentioned that. I am saying that 100 m would not be the ground level .... At 100 m ball would still be in the air because u started from a height.

I think Erythropoietin is saying that before launch, KE = 0, PE = 1000. Then when the balls are descending and once again reach 100m above the ground (same height as starting point), PE = 1000 and KE = 0. Then, just before the balls hit the ground, PE = 0 and KE = 1000. So TE is the same start and finish.

Is that right?

 

[Erythropoietin]

I think Erythropoietin is saying that before launch, KE = 0, PE = 1000. Then when the balls are descending and once again reach 100m above the ground (same height as starting point), PE = 1000 and KE = 0. Then, just before the balls hit the ground, PE = 0 and KE = 1000. So TE is the same start and finish.

Is that right?

Yes, what I said was quite confusing. When you are at the cliff at 100 m and not moving, you have 1000 J of potential energy. But since you are projecting it upward "from a cliff", I treated the 100 m level as ground level, and only cared about the projectile motion and ignored whatever is below the cliff, hence "ground level".
Since you are projecting it upwards by some degree from the ground, you NEED some initial velocity, right? So I was saying that velocity comes from the 1000 J of kinetic energy from the beginning of the projectile. When the projectile goes to its max height and comes back down to ground level (100 m), you will still have 1000 J of energy because of energy conservation. Since the ball is not going to stop in the air, it will fall down 100 m WITH 1000 J of energy (because that's how it initially started in the first case).
So I'm saying that a ball that is projected only horizontally will still have the same final kinetic energy with a ball that is projected upward from a cliff which then reaches the ground because they all start with the same initial potential energy.

The only difference will be in their displacement from the cliff (since they are launched at different angles) but the question isn't asking that.

Hope that makes it more clear

 

[Pisiform]

Why u are saying you are not moving. You have a velocity there.

 

[Erythropoietin]

Why u are saying you are not moving. You have a velocity there.

I'm saying at the very beginning when you didn't projected it upwards yet, you are not moving.

 

[Pisiform]

I'm saying at the very beginning when you didn't projected it upwards yet, you are not moving.

Yea but did they say they started from rest?

 

[Erythropoietin]

Yea but did they say they started from rest?

I was giving an example to explain my point.

 

[MedPR]

Yea but did they say they started from rest?

In the problem, the balls must be at rest before they are shot out of the cannon.