KE realated to translational motion

[helpmeplzzz]

The question describes a scenario in which three balls of equal mass are launched from a cannon on top of a 100m cliff at the same speed (50 m/s) but at different angles (0, 30, and 60 degrees). The question asks which of the following balls hits the ground with the greatest kinetic energy. The answer is that all of them hit the ground with the same kinetic energy. Their reasoning is that all the balls have the same initial and final PE and same initial KE, as they all start from the same height and end at the same height.

(PE + KE)_final = (PE + KE)_initial

I'm confused because intuitively this doesn't make sense to me. Even though they start at the same height and finish at the same height, the balls don't hit the ground with the same velocity, which should affect their final KE based on the equation KE = 1/2 mv². No?

 

[berkeman]

The question describes a scenario in which three balls of equal mass are launched from a cannon at the same speed (50 m/s) but at different angles (0, 30, and 60 degrees). The question asks which of the following balls hits the ground with the greatest kinetic energy. The answer is that all of them hit the ground with the same kinetic energy. Their reasoning is that all the balls have the same initial and final PE and same initial KE, as they all start from the same height and end at the same height.

(PE + KE)_final = (PE + KE)_initial

I'm confused because intuitively this doesn't make sense to me. Even though they start at the same height and finish at the same height, the balls don't hit the ground with the same velocity, which should affect their final KE based on the equation KE = 1/2 mv². No?

Yeah, the wording seems suspicious. Can you post the text of the exact question and scan any diagram that goes with the question?

 

[oli4]

They won't follow the same path, some will go farther, others higher, but the velocity although differently oriented as a vector will have the same norm. the speed will be the same, and the K.E will be the same

 

[LawrenceC]

I have deleted my post.

 

[5-Sigma]

If you think about it, a ball fired vertically upwards doesn't back travel down with additional energy and lands with the same velocity as it had when it was originally fired (opposite direction) assuming no air resistance. With this in mind, its easy to see why they both have the same kinetic energy when they land. If they have the same mass then they must land with the same speed of course.

So why does the 60 degree ball intuitively (perhaps) seem to be traveling faster? Well in the case of the 60degree ball, the vast majority of its velocity will be in the vertical component, the 0Degree ball will have a greater horizontal component. By adding the two components together, you will get the same vmax but in different directions. The ball with the higher vertical velocity will have the bigger impact on landing, this is why it might intuitively seem to travel faster overall.

 

[helpmeplzzz]

If you think about it, a ball fired vertically upwards doesn't back travel down with additional energy and lands with the same velocity as it had when it was originally fired (opposite direction) assuming no air resistance. With this in mind, its easy to see why they both have the same kinetic energy when they land. If they have the same mass then they must land with the same speed of course.

So why does the 60 degree ball intuitively (perhaps) seem to be traveling faster? Well in the case of the 60degree ball, the vast majority of its velocity will be in the vertical component, the 0Degree ball will have a greater horizontal component. By adding the two components together, you will get the same vmax but in different directions. The ball with the higher vertical velocity will have the bigger impact on landing, this is why it might intuitively seem to travel faster overall.

OK, I think that's where my confusion was. Thanks for clearing it up.