Question

You are in a hot air balloon (yes, another balloon problem!) rising from the ground at a constant velocity of 2.00 m/s upward. to celebrate the takeoff, you open a bottle of champagne, expelling the cork with a horizontal velocity of 6.60 m/s relative to the balloon. when opened, the bottle is 5.90 m above the ground. what is the initial speed of the cork, as seen by your friend on the ground?

Answer 1

If all you need is the initial speed of the cork, you can solve this using only two of your given:
2.00 m.s upward and 6.60 m.s horizontally.

If you take in consideration the movement of the cork, you know that it was both going up and forward at the same time, this means that it was moving at a diagonal direction. Now you can solve this by using the Pythagorean theorem where: 

$c = \sqrt{ a^{2} + b^{2} }$

Why? Because the vertical and the horizontal motion creates a movement that is diagonal, which when put in a free-body diagram, creates a right triangle. 

Going back to your problem, when applying this, the diagonal of a right triangle is the hypotenuse, so this is what you are looking for. The horizontal and vertical motion will represent the other 2 sides of the triangle. 

Now let's put that into your formula:

$c = \sqrt{ a^{2} + b^{2} }$

$Vi = \sqrt{ Vx^{2} + Vy^{2} }$

Where: Vx is your horizontal velocity
             Vy is your vertical velocity
             Vi  is your initial velocity

Now let's put in your given:

$Vi = \sqrt{ Vx^{2} + Vy^{2} }$
$Vi = \sqrt{ 6.60^2} + 2.00^{2} }$
$Vi = \sqrt{ 43.56 + 4.00 }$
$Vi = \sqrt{ 47.56 }$
$Vi = 6.8964 m/s$

So your initial velocity is 6.8964 m/s or 6.90 m/s