Motions with zero launch angle.

Click For Summary

Homework Help Overview

The problem involves a hot-air balloon rising vertically while a cork is expelled horizontally from a champagne bottle. The task is to determine the initial velocity and speed of the cork as observed from the ground. The context includes elements of kinematics and vector addition.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial velocities of the cork in both horizontal and vertical directions, with some confirming values and others questioning the equations related to maximum height.

Discussion Status

Some participants have provided confirmations regarding the initial velocities, while others are exploring different equations related to projectile motion. There is an ongoing examination of the relationships between the variables involved.

Contextual Notes

Participants are navigating potential confusion regarding the definitions of variables and the application of kinematic equations, particularly in the context of projectile motion from a moving reference frame.

matt@USA
Messages
25
Reaction score
0

Homework Statement


A hot-air balloon rises from the ground with a velocity of (3.00m/s)y. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (4.00m/s)x relative to the balloon. When opened, the bottle is 5.00m above the ground.

What is the initial velocity of the cork, as seen by an observer on the ground?

What is the speed of the cork, as seen by the same observer?



Homework Equations


I am just confused as to what is what. All I need help with is, is making sure I am correct. I would say y=5.0m, and x=0 correct? I would then say that Vnotx is 4.0m/s, and Vnoty is 3.0m/s?



The Attempt at a Solution

 
Physics news on Phys.org
Someone please help!
 
I would then say that Vnotx is 4.0m/s, and Vnoty is 3.0m/s?
That is correct for part 1. By the way it's V-naught-x/y, or better: V0x, V0y. :smile:
 
Is ymax not Vnaught^2(sin)^2theta/2(g)?
 
I prefer the more readable and less ambiguous: ymax = (v02sin2Θ)/2g.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
25
Views
3K
Replies
11
Views
3K
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
11K
Replies
8
Views
10K
  • · Replies 3 ·
Replies
3
Views
9K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K