Motions with zero launch angle.

AI Thread Summary
The discussion focuses on calculating the initial velocity of a cork expelled from a champagne bottle opened in a hot-air balloon. The balloon rises at 3.00 m/s vertically, while the cork is expelled horizontally at 4.00 m/s relative to the balloon. The initial velocities are confirmed as V0x = 4.00 m/s and V0y = 3.00 m/s. Participants clarify the notation for initial velocities and discuss the formula for maximum height in projectile motion. The conversation emphasizes understanding the components of velocity in different reference frames.
matt@USA
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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

 
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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.
 
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