Motion in two or three dimensions

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Homework Help Overview

The discussion revolves around a problem related to motion in two or three dimensions, specifically involving the analysis of vertical and horizontal components of velocity in a projectile motion context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants express confusion about how to begin solving the problem and inquire about the appropriate equations to use. There are discussions about breaking down the initial speed into vertical and horizontal components, with specific references to trigonometric functions.

Discussion Status

Some participants have provided guidance on considering the components of velocity and have suggested equations to use. There is an ongoing exploration of the relationships between the components and the time taken for different parts of the motion. However, there is no explicit consensus on the correctness of the equations being discussed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can access or the methods they can employ. There is mention of gravitational effects in the equations, indicating that the problem involves vertical motion influenced by gravity.

Faka
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Hi. I am lost in this problem. Could someone tell me what to do. How to start with the problem?
 

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Faka said:
Hi. I am lost in this problem. Could someone tell me what to do. How to start with the problem?

Start by considering the vertical and horizontal components of the initial speed.
 
Which equation do I need to use?
 
Faka said:
Which equation do I need to use?

One of the components is Vsin(53) the other is Vcos(53)
 
Vx = V0*cos(53,0)

Vy = V0*sin(53,0)
 
Faka said:
Vx = V0*cos(53,0)

Vy = V0*sin(53,0)

Now you have to use them.

One will tell you how long it takes to get across the river, the other will let you calculate how long it takes the car to go up in the iar, then come back down again to the height of the far bank, and those two times have to be the same.
 
I have just seen in my book, that the last equations needs to be:

Vy = V0*sin(53,0) - g*t

Is that true?

So I get these two equation, that I need to use as you said:

Vx = V0*cos(53,0)

Vy = V0*sin(53,0) - g*t
 

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