The shell strikes the water at the same speed?

[PainterGuy]

Hi

Please have a look on the attachment. Do you also agree that the shell strikes the water at the same speed?

It seems false to me.

The vertical and horizontal motion is independent of each other. When the shell is fired horizontally, it starts descending toward the ground from less height compared to the other case. It's kinetic energy is used to push the shell horizontally against air resistance.

On the other hand, when the shell is fired at an angle, it's kinetic energy has two components. One component pushes it forward horizontally and the other component pushes it vertically against the gravitational force . In this case the shell descends toward the ground from greater height and hence more kinetic energy at the moment when it strikes water. Thank you.

 

[Doc Al]

Do you also agree that the shell strikes the water at the same speed?

Yes. (Forget about air resistance.)

When the shell is fired horizontally, it starts descending toward the ground from less height compared to the other case.

So? The total energy it has is the same regardless of its firing direction. That's what will determine the speed when it hits the ground.

If shot at an upward angle, when it falls back to the original height it will have the same speed that it started with.

 

[PainterGuy]

I'm sorry but it still doesn't make sense to me.

Let me try. It looks like that I was interpreting it wrongly.

Case 1: Suppose that the 0.1 kg shell has 10 J kinetic energy when it is fired horizontally and it's height from water is 5 m. Ignoring air resistance, at the moment of coming in touch with water, it will possesses kinetic energy (10 J) and kinetic energy it gained thru its free fall (0.1x10x5 = 5J); total 15 J. Its speed will have two components - horizontal and vertically downward.

Case 2: When it is fired at an angle, its energy consists of kinetic energy which will have two component - one pushing it vertically upward (5 J) and the other pushing it horizontally (5 J). After reaching a certain height, say, 5 m above the surface where gun is placed, all of the component of its kinetic energy which was pushing it upward is converted into potential energy, and at that moment its potential energy would be 10 J. The other component which pushes the shell horizontally would be intact. At the moment of striking the water, it will possesses kinetic energy (5 J) and kinetic energy it gained thru its free fall (0.1x10x10 = 10 J); total 15 J. Again, at the strike its speed will have two components.

If you ask me, the comparison between the two cases is misleading because the resultant speed at the moment of strike in both cases will be different. Thank you.

 

[sophiecentaur]

Let me try. It looks like that I was interpreting it wrongly.

Yep. You were.
Forget the complicated stuff about forces and directions; an argument involving Energy is the clincher. The Energy when it hits the water is the same, whatever path it takes and it's equal to the Kinetic Energy from the gun and the Gravitational Potential Energy at the launch height. All that energy turns up as Kinetic Energy of the shell entering the water. (Independent of launch angle).
If you want to, you can go through the sums, going through your alternative route, but you can't hope to get a different answer. And you have to actually do the complete sums - not the half calculations you did in your post.
Energy is your friend in all calculations like this.

 

[Quandry]

I'm sorry but it still doesn't make sense to me.

Let me try. It looks like that I was interpreting it wrongly.

Case 1: Suppose that the 0.1 kg shell has 10 J kinetic energy when it is fired horizontally and it's height from water is 5 m. Ignoring air resistance, at the moment of coming in touch with water, it will possesses kinetic energy (10 J) and kinetic energy it gained thru its free fall (0.1x10x5 = 5J); total 15 J. Its speed will have two components - horizontal and vertically downward.

Case 2: When it is fired at an angle, its energy consists of kinetic energy which will have two component - one pushing it vertically upward (5 J) and the other pushing it horizontally (5 J). After reaching a certain height, say, 5 m above the surface where gun is placed, all of the component of its kinetic energy which was pushing it upward is converted into potential energy, and at that moment its potential energy would be 10 J. The other component which pushes the shell horizontally would be intact. At the moment of striking the water, it will possesses kinetic energy (5 J) and kinetic energy it gained thru its free fall (0.1x10x10 = 10 J); total 15 J. Again, at the strike its speed will have two components.

If you ask me, the comparison between the two cases is misleading because the resultant speed at the moment of strike in both cases will be different. Thank you.

Two factors here. The shell starts to fall at the same rate from the time it leaves the barrel. However, depending on the angle that it is fired at, it may still have some of the energy imparted to it by firing when it hits the ground. e.g. If you fire it straight up in the air it will rise until all of the imparted energy is lost to the extent that gravity takes over and it starts to descend. Then the speed with which it hits the ground will be 1/2gt² where t is dependent on the height it reached.
Alternately, if the shell is fired directly at the ground, the shell will hit with its original velocity, plus gravity, less ballistic losses.
Between these two extremes the shell will strike with a velocity which has a component of gravity and original velocity. So the speed that it 'hits the water's your examples will always be different.

 

[Doc Al]

Case 2: When it is fired at an angle, its energy consists of kinetic energy which will have two component - one pushing it vertically upward (5 J) and the other pushing it horizontally (5 J).

Velocity has components; energy does not. But I understand what you mean: You can break the velocity into vertical and horizontal components and calculate the KE associated with each component to get the total.

After reaching a certain height, say, 5 m above the surface where gun is placed, all of the component of its kinetic energy which was pushing it upward is converted into potential energy, and at that moment its potential energy would be 10 J.

Here's your error. If the vertical component of velocity only accounts for half the total KE then the highest it can go would be where its potential energy would be 5 J. So the total energy at any point is always be 10 J.

 

[Doc Al]

So the speed that it 'hits the water's your examples will always be different.

Not so. (Keep to the simpler case and ignore air resistance.) Whether you fire the bullet straight down into the water or straight up into the air, when it hits the water it will have the same speed.

 

[Quandry]

Not so. (Keep to the simpler case and ignore air resistance.) Whether you fire the bullet straight down into the water or straight up into the air, when it hits the water it will have the same speed.

Absolutely not the case. A projectile fired vertically will loose its initial velocity and start to fall. Assuming that it takes 10 seconds to do so, it will have a velocity of 980 fps when it hit the water. If fired directly at the water at a typical velocity of 2500 fps it will hit the water at 2500 fps.

 

[Doc Al]

A projectile fired vertically will loose its initial velocity and start to fall.

A projectile fired vertically upward will lose velocity as it rises -- then gain it back as it falls back down. Energy is conserved. (Ignoring the complications of air resistance.)

 

[Nugatory]

Absolutely not the case. A projectile fired vertically will loose its initial velocity and start to fall. Assuming that it takes 10 seconds to do so, it will have a velocity of 980 fps when it hit the water. If fired directly at the water at a typical velocity of 2500 fps it will hit the water at 2500 fps.

If the speed at which the shell leaves the barrel is 2500 fps then if fired straight up it will reach a height such that it takes 2500/32 seconds to fall back down, and will be moving downwards with a speed of 2500 fps, just as if it had been fired straight down in the first place.

 

[sophiecentaur]

Absolutely not the case.

Your argument would only hold if Energy were not Conserved. It is conserved so you just have to be wrong - without any calculations needed. You need to find out what's actually wrong about your argument, rather than try to prove that it's right. Do the complete calculation and, as other posts have shown, the speeds will be the same.
(You have not found a 'flaw in Physics'.)

 

[Quandry]

Your argument would only hold if Energy were not Conserved. It is conserved so you just have to be wrong - without any calculations needed. You need to find out what's actually wrong about your argument, rather than try to prove that it's right. Do the complete calculation and, as other posts have shown, the speeds will be the same.
(You have not found a 'flaw in Physics'.)

Your argument is only valid if you ignore the conversion of energy relating to the losses. Do the Complete calculation.

 

[sophiecentaur]

Your argument is only valid if you ignore the conversion of energy relating to the losses. Do the Complete calculation.

Of course. The length of path through the air will naturally affect the drag losses. But so far, that has explicitly been ignored because it is not fundamental to the OP.
Get one thing right at a time and that means you need to do the complete calculation, ignoring losses first.

 

[Quandry]

If the speed at which the shell leaves the barrel is 2500 fps then if fired straight up it will reach a height such that it takes 2500/32 seconds to fall back down, and will be moving downwards with a speed of 2500 fps, just as if it had been fired straight down in the first place.

You would need to explain the logic of that statement.

 

[sophiecentaur]

and will be moving downwards with a speed of 2500 fps,

He missed out "when it passes the launch position" (but we 'all' realized that). It will obviously accelerate further from that position to the water surface.
But it all goes to show that trying to 'prove' something using magic numbers can lead to confusion and also that avoiding the Energy Argument is usually not the best way.
@Quandry : are you arguing against the simple Energy-based explanation?

 

[Nugatory]

Some posts about the effects of drag and air resistance have been removed from thus thread. The original question is complicated enough without introducing that complication.