That's what I've always thought.

However, I just got a question asking me if a 4kg projectile has 1400 J of kinetic energy when it leaves the ground, and 120 Joules of kinetic energy at its highest point, how high did it travel?

Apparently, we can do 1400-120=1280J and just throw the 1280 in the u=mgh equation.

However, isn't KE always zero at the highest point since velocity = 0 ? I always thought that at the highest point, all of the kinetic energy would have been converted to potential energy?

any help would be appreciated. thanks:luck:

You are right to an extent....
Whenever you think of projectiles, split it up into the x and y direction. So, at the highest point, there is no velocity in the y direction, but there is still velocity in the x direction. Since KE is nt a vector, the KE would just be 1/2mv^2 using v in the x direction. (Whenever there is velocity, there must be KE). However, if you look at the y-direc alone, you see that all the KE in that direction was transferred to PE. v(y-direc)=sqrt(2gh). So in the y-direc, all the KE was transferred into PE.
Overall, the highest point in a projectile has both KE and PE. However, if you were to throw an object straight upward, it would have no velocity in the x-direc and would therefore have no KE at it's highest point.
Hope this helps a little...:luck:

jg's right =).

This is how any projectile works though, and why it is generally better to solve projectile problems using basic kinematics instead of energy equations (unless you have a firm grasp on the concepts or can't solve it another way). Any projectile shot at an angle other than 0 or 90 will have an x component and a y component.

To throw some further completely unnecessary info that you may find useful anyways:

It may be confusing to think that energy is a scalar, because it seems as though here we have energy in multiple directions (if you have a velocity component in an x and y direction, why can't you have a kinetic energy component in x and y directions).

What it means to say kinetic energy is a scalar is that the kinetic energy of a system is simply the sum of its individual kinetic energies, regardless of the direction of the velocity. This is different to velocity in which we would need to use vector addition.

jg's right =).

This is how any projectile works though, and why it is generally better to solve projectile problems using basic kinematics instead of energy equations (unless you have a firm grasp on the concepts or can't solve it another way). Any projectile shot at an angle other than 0 or 90 will have an x component and a y component.

To throw some further completely unnecessary info that you may find useful anyways:

It may be confusing to think that energy is a scalar, because it seems as though here we have energy in multiple directions (if you have a velocity component in an x and y direction, why can't you have a kinetic energy component in x and y directions).

What it means to say kinetic energy is a scalar is that the kinetic energy of a system is simply the sum of its individual kinetic energies, regardless of the direction of the velocity. This is different to velocity in which we would need to use vector addition.

You are right to an extent....
Whenever you think of projectiles, split it up into the x and y direction. So, at the highest point, there is no velocity in the y direction, but there is still velocity in the x direction. Since KE is nt a vector, the KE would just be 1/2mv^2 using v in the x direction. (Whenever there is velocity, there must be KE). However, if you look at the y-direc alone, you see that all the KE in that direction was transferred to PE. v(y-direc)=sqrt(2gh). So in the y-direc, all the KE was transferred into PE.
Overall, the highest point in a projectile has both KE and PE. However, if you were to throw an object straight upward, it would have no velocity in the x-direc and would therefore have no KE at it's highest point.
Hope this helps a little...:luck:

Thank you both very much. It makes a lot of sense now.

However, in that case, why can't we use that initial 1400J of kinetic energy, and simply find the initial velocity from it in KE=.5mv^2? Wouldn't "initial energy" mean "initial velocity" ?

since we still have KE at the highest point, then we can state with certainty that the object was thrown at some angle. If the question would have given us the angle the object was thrown at, could we have taken that initial energy, used it to find the initial velocity, and used the initial velocity to find the initial velocity of the Y axis?

Is that why we can't use that initial energy to get the initial velocity and simply put it in v^2=vo^2+2ax ?

again, thank you both:luck:

Let me see if I can help you with this one. What the other posters have said is dead on, but let me elaborate.

You should split this problem up and think about it in two directions. Think about the motion in the y direction and the motion in the x direction. Okay, so you're given that the total KE is 1400J, and the KE at the highest point is 120J.
Since the y-velocity is zero at the highest point, 120J is the KE for horizontal motion. Now, think about the KE due to horizontal motion at any point during the flight of the object.
You know that:
1) KE is 1/2mv^2,
2) and you know that the horizontal velocity of a projectile object never changes. That is, the object moves horizontally at the same speed at any time in flight.

So if the horizontal speed is the same at any time in flight, and the mass is constant, then the KE contribution from horizontal velocity is the same at any point in time.

So what does this mean? This means that since the KE at the top of flight, is the same as the KE contribution from the x-velocity at the begining, 120J from that original 1400J must be due to horizontal motion.

That means that the remainder of the energy must be driving the ball in the vertical direction. So that means that 1400-120 = 1280J is the amount of energy that is driving the object in the y-direction.

In effect, if you realize that the KE from horizontal motion is the same at any point, you can simplify this problem into a much simpler problem:

"If an object is thrown directly upward and its initial KE is 1280J, what is the maximum height it reaches?"

And then since you know that there is no PE at the bottom and no KE at the top, because energy is conserved that means that initial KE = apex PE.
Thus, 1280 = mgh. And you can solve for h.

So the trick is seeing that you can separate the problem into two directions, and thus you have a much simpler one-dimensional problem. GL :)

Thank you both very much. It makes a lot of sense now.

However, in that case, why can't we use that initial 1400J of kinetic energy, and simply find the initial velocity from it in KE=.5mv^2? Wouldn't "initial energy" mean "initial velocity" ?

since we still have KE at the highest point, then we can state with certainty that the object was thrown at some angle. If the question would have given us the angle the object was thrown at, could we have taken that initial energy, used it to find the initial velocity, and used the initial velocity to find the initial velocity of the Y axis?

Is that why we can't use that initial energy to get the initial velocity and simply put it in v^2=vo^2+2ax ?

again, thank you both:luck:

Yea this would be possible. Immediately after the projectile is fired the energy is in the form of kinetic energy. If you have the initial KE, you can find the velocity from KE = .5 m v^2. Now note this is the total velocity not the velocity components. As a result, this will give the total KE. Alternatively if you were to take each directional component and find their KE's and then sum both values up you would get the same value.

As a working example:

Lets say you have a velocity of 5m/s, an x velocity of 3, and a y velocity of 4 (3 4 5 triangle). You have a mass of 1kg.

KE = .5 * 1 * 5^2 = 12.5
KEx = .5 * 1 * 3^2 = 4.5
KEy = .5 * 1 * 4^2 = 8

KEx + KEy = 12.5 = KE

As you can see, the sum of the kinetic energy contributed from each directional component is equal to the total kinetic energy. Also note this is a scalar because it's a direct sum to get the total (as oppose to velocity, which is a vector, where the x component and the y component don't sum directly)

So to answer your question, if you know the initial KE you can find the initial v. If you know the angle you can find the vertical and horizontal velocity components.

Let me see if I can help you with this one. What the other posters have said is dead on, but let me elaborate.

You should split this problem up and think about it in two directions. Think about the motion in the y direction and the motion in the x direction. Okay, so you're given that the total KE is 1400J, and the KE at the highest point is 120J.
Since the y-velocity is zero at the highest point, 120J is the KE for horizontal motion. Now, think about the KE due to horizontal motion at any point during the flight of the object.
You know that:
1) KE is 1/2mv^2,
2) and you know that the horizontal velocity of a projectile object never changes. That is, the object moves horizontally at the same speed at any time in flight.

So if the horizontal speed is the same at any time in flight, and the mass is constant, then the KE contribution from horizontal velocity is the same at any point in time.

So what does this mean? This means that since the KE at the top of flight, is the same as the KE contribution from the x-velocity at the begining, 120J from that original 1400J must be due to horizontal motion.

That means that the remainder of the energy must be driving the ball in the vertical direction. So that means that 1400-120 = 1280J is the amount of energy that is driving the object in the y-direction.

In effect, if you realize that the KE from horizontal motion is the same at any point, you can simplify this problem into a much simpler problem:

"If an object is thrown directly upward and its initial KE is 1280J, what is the maximum height it reaches?"

And then since you know that there is no PE at the bottom and no KE at the top, because energy is conserved that means that initial KE = apex PE.
Thus, 1280 = mgh. And you can solve for h.

So the trick is seeing that you can separate the problem into two directions, and thus you have a much simpler one-dimensional problem. GL :)

Yea this would be possible. Immediately after the projectile is fired the energy is in the form of kinetic energy. If you have the initial KE, you can find the velocity from KE = .5 m v^2. Now note this is the total velocity not the velocity components. As a result, this will give the total KE. Alternatively if you were to take each directional component and find their KE's and then sum both values up you would get the same value.

As a working example:

Lets say you have a velocity of 5m/s, an x velocity of 3, and a y velocity of 4 (3 4 5 triangle). You have a mass of 1kg.

KE = .5 * 1 * 5^2 = 12.5
KEx = .5 * 1 * 3^2 = 4.5
KEy = .5 * 1 * 4^2 = 8

KEx + KEy = 12.5 = KE

As you can see, the sum of the kinetic energy contributed from each directional component is equal to the total kinetic energy. Also note this is a scalar because it's a direct sum to get the total (as oppose to velocity, which is a vector, where the x component and the y component don't sum directly)

So to answer your question, if you know the initial KE you can find the initial v. If you know the angle you can find the vertical and horizontal velocity components.


thank you both

you have no idea how much that helped. Thank you. I am having such a horrible time with physics. I wish every explanation were like this

Quick question: since u=mgh, should we assume that horizontal potential energy for projectiles is always zero?

:luck:

thank you both

you have no idea how much that helped. Thank you. I am having such a horrible time with physics. I wish every explanation were like this

Quick question: since u=mgh, should we assume that horizontal potential energy for projectiles is always zero?

:luck:

u = mgh is gravitational potential energy and is always along the y axis because that's the direction gravity acts. If there is a horizontal potential energy in a projectile system, it won't be due to gravity so I think it's a safe assumption to make. Bear in mind though that you can have a horizontal potential energy (such as a spring laying horizontally). Know that it's always easier to fully understand the concept than to memorize possible assumptions =).