Velocity on a moving reference frame

[Calpalned]

Homework Statement

I am reviewing for my midterm and I just want to quickly understand a concept. It seems that velocity does not depend on the original reference frame. For example, let's say spider man is standing on the ground next to a 78.4 meter tall building, and it is known that if he shoots a web, it will reach the top of the building in 3 seconds. Now let's assume that he falls off of the building and shoots the web after falling for one second. It takes a total of four seconds for spiderman to land. According to the solutions to the practice midterm, the web will still reach the roof in three seconds. I don't understand this. First of all, why is the distance that is covered 78.4 meters, and not the displacement of spiderman after one second? Secondly, wouldn't it take longer for the web to reach the ledge, given that SpiderMan is initially moving downward?

Homework Equations

N/A

The Attempt at a Solution

 

[BvU]

Well now, anything to stay within the PF rules, eh ?
I don't see a problem statement, and I don't see your attempt at solution.

 

[nrqed]

View attachment 83326

Homework Statement

I am reviewing for my midterm and I just want to quickly understand a concept. It seems that velocity does not depend on the original reference frame. For example, let's say spider man is standing on the ground next to a 78.4 meter tall building, and it is known that if he shoots a web, it will reach the top of the building in 3 seconds. Now let's assume that he falls off of the building and shoots the web after falling for one second. It takes a total of four seconds for spiderman to land. According to the solutions to the practice midterm, the web will still reach the roof in three seconds. I don't understand this. First of all, why is the distance that is covered 78.4 meters, and not the displacement of spiderman after one second? Secondly, wouldn't it take longer for the web to reach the ledge, given that SpiderMan is initially moving downward?

Homework Equations

N/A

The Attempt at a Solution

View attachment 83326

First, note that when they give the speed of the web, they are picking a reference frame with respect to which the speed is measured. It looks like they are calculating the speed from the frame of Spidey and it is obvious that this must be the same no matter if he is falling or not, if we look at the situation from the POV of Spidey. But it can be tricky to look at a situation from different POVs and it can be a bit tricky to switch back and forth. The first point that is crucial to keep in mind is that distances traveled are different in different frames, and velocities also depend on the frame of reference. But the time interval is the same (in classical mechanics, things are quite different in special relativity!).

To get to your question, when Spidey is falling, the distance covered by the web depends on the POV. From the POV of Spidey, here is what is happening: He shoots the web when the top of the building is already a certain distance above him. He then sees the top of the building moving upward from him as he falls. At the instant he lands on the ground, he sees the web reaching the top of the building, 78.4 m above. So clearly, from his POV, the web traveled the height of the building.

Now, from the POV of someone on the ground (not falling with Spidey), the web does NOT travel the entire height of the building. You are right about this! The web only travels the distance Spiderman was below the top of the building at the instant he shot the web. On the other hand, the speed of the web is much less from that point of view. It takes again 3 seconds for the web to reach the top of the building even though the distance traveled is much less than 78.4 m simply because the speed in that frame is less than 26.1 m/s.

Hope this helps