Projectile motion and time [SOLVED]

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Homework Help Overview

The problem involves a hot air balloon rising and the subsequent horizontal launch of a cork from a champagne bottle. The cork's initial vertical velocity and height above the ground are provided, and the question seeks to determine the total time the cork remains in the air.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the vertical motion of the cork and suggest focusing on the vertical component of motion to determine the time of flight. There is mention of using kinematic equations to analyze the motion.

Discussion Status

Some participants have provided guidance on which equations to consider for solving the problem, while others are exploring different interpretations of the motion involved. There is acknowledgment of previous attempts and some uncertainty regarding calculations.

Contextual Notes

Participants note the need to account for both the upward motion and the subsequent descent of the cork, indicating a focus on the effects of gravity and initial conditions. There is mention of potential complications in calculations and the use of the quadratic formula.

annersmarie
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[SOLVED] Projectile motion and time

Homework Statement



A hot air balloon rises from the ground with a velocity of (2.00 m/s) in the y direction. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (5.00 m/s) in the x direction, relative to the balloon. When opened, the bottle is 6.00 m above the ground.

How long does the cork remain in the air?


Homework Equations



There are other parts of this problem that I've already solved. For example, I've found that the initial velocity and direction of the cork, as seen by an observer on the ground, is 5.39 m/s at 21.8 degrees above horizontal (w/ respect to the balloon) with these equations:
V= [(5.0)^2+(2.0)^2]^1/2

theta= arctan (2.0/5.0)

I've also found that the maximum height above the ground attained by the cork is 6.204 meters by finding:

h= [(5.39^2)sin^2(21.8)]/2(9.81)

and adding it to the initial height of 6.00 m.

The Attempt at a Solution



I know that the equation to find the time that a projectile is in the air is as follows:

t= [2Vo*sin(theta)]/g

but I know that that only accounts for the time that it will take the cork to reach a height of 6 meters. What I don't know is how to account for the rest of the time.
 
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Forget about the corks velocity in the x direction, all you need to know is the vertical. The question tells you that the cork is moving 2 m/s upward and starts off 6m above the ground.

From there you can use constant acceleration formula to determine what the maximum height of the cork is and how long it takes to get there, and then how long it takes to fall to the ground again.
 
Welcome to PF,

I'm not checking your math, but your method looks good to me. For the time of flight, forget your time equation there and consider one of the general kinematic equations (found https://www.physicsforums.com/showpost.php?p=905663&postcount=2"). You know the initial displacement, the initial velocity as well as the final displacement, so which equation do you think you should use?
 
Last edited by a moderator:
I guess it would be the displacement and time equation with uniform acceleration, just in the y dimension instead of the x. I'll get back with you after I do the math to see if I've got this figured out.

Thanks so far, guys.
 
Finished

Awesome. It was that equation after all. I actually tried it before I posted the first time, but I think my math was off somewhere when I used the quadratic formula to solve for t. Then I just ended up overcomplicating things for myself. Sometimes you're your own worst enemy, y'know?

Thanks again for the help. I really appreciate it.

-Anne
 
annersmarie said:
Awesome. It was that equation after all. I actually tried it before I posted the first time, but I think my math was off somewhere when I used the quadratic formula to solve for t. Then I just ended up overcomplicating things for myself. Sometimes you're your own worst enemy, y'know?
I know the feeling :rolleyes:
annersmarie said:
Thanks again for the help. I really appreciate it.

-Anne
A pleasure :smile:
 

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