Solving a Hot Air Balloon & Champagne Cork Problem

[baseketballr09]

Homework Statement

A hot air balloon rises from the ground with a velocity of (2.0m/s)y. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (5.0m/s)x. When opened, the bottle is 6 m above the ground. a) What is the initial velocity of the cork, as seen by an observer on the ground? What is the speed of the cork, and its initial direction of motion, as seen by the same observer? b) Determine the maximum height above the ground attained by the cork. c) how long does the cork remain in the air?

Homework Equations

The Attempt at a Solution

you guys I'm so totally lost, please please please help me out. ILL PAY YOU. Add my aim sn: baseketballr09 or my msn: bballrkhan@hotmail.com and i'll get you $5-10 via paypal, honestly. This is so important, it determines the difference between a 3.7 gpa and a 2.9 gpa, because i have a huge quiz tomorrow and this is one of the problems that will be on it. THIS IS REALLY URGENT, IF YOU EVER GAVE A HOOT ABOUT YOUR GRADE THEN I HOPE YOU FEEL FOR ME, I'M REALLY STRESSED OUT AND I WOULD MUCH APPRECIATE HELP...I'M BEGGING YOU GUYS. JUST HELP ME OUT THIS ONCE :(

 

[Shooting Star]

Relax. You have to answer at least one question before I solve it for you.

What was the initial speed of the cork when it was expelled from the bottle, as seen by the ground observer?

 

[baseketballr09]

if initial velocity and speed are the same thing...then i got 5.4 m/s

 

[Shooting Star]

That’s right. Of course, I don’t know how you got it. Speed is the magnitude of velo.

a) The initial velo of the cork as seen from the ground has two components, vx and vy.
vx = 5 m/s and vy = 2 m/s. So, initial speed = sqrt(vx^2+vy^2)= 5.38 m/s. The velocity is given by the 2 components, and if you take x–axis in the horizontal dirn in which the cork is travelling, and y-axis in the upward dirn, then you can write initial vi = 2 i + 5 j.
If theta is the angle vi makes with the horizontal, then tan(theta)= vy/vx = 2/5. That gives you the dirn of motion as seen from the ground.

b) If it rises h m above its starting position, then vy^2=2gh => h= 0.2 m. So, it’ll rise 6.2 m above the ground.

c) If t secs is the time it stays in the air, then taking the origin at its starting point we have,
-6 = vy*t – gt^2/2 => -6 = 2t – 9.8*t^2/2. This is a quad eqn in t. You can solve it and take the +ve root.