A hot-air balloon rises from the ground with a velocity of (2.00 m/s)y. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (6.00 m/s)x relative to the balloon. When opened, the bottle is 7.00 m above the ground.

1)What is its initial direction of motion, as seen by the same observer?
2)Determine the maximum height above the ground attained by the cork
3)How long does the cork remain in the air?

1) The initial direction of motion of the cork, as seen by the same observer, is horizontal. It is moving in the positive x direction.

2) To determine the maximum height above the ground attained by the cork, we can use the equations of motion. We'll assume that the only force acting on the cork is gravity, neglecting air resistance.

The vertical motion of the cork can be described by the equation:

y = y₀ + v₀yt - 0.5gt²

where:
- y is the displacement (height above the ground)
- y₀ is the initial displacement (7.00 m)
- v₀y is the initial vertical velocity (0 m/s, since the cork is launched horizontally)
- g is the acceleration due to gravity (-9.8 m/s²)
- t is the time

Since the initial vertical velocity is 0 m/s, the equation simplifies to:

y = y₀ - 0.5gt²

At maximum height, the vertical velocity of the cork will be zero. So, we can find the time it takes for the cork to reach maximum height by setting vₓ = 0:

0 = v₀y - gt

Solving for t:

t = v₀y / g

Plugging in the values, we have:

t = 0 / (-9.8) = 0

This means that the cork reaches maximum height instantaneously. Therefore, the maximum height above the ground attained by the cork is the initial height of 7.00 m.

3) Since the cork reaches maximum height instantaneously before falling back to the ground, it remains in the air for a very short duration of time.

To answer these questions, we need to analyze the motion of the hot-air balloon and the cork separately.

1) What is the initial direction of motion of the hot-air balloon, as seen by the same observer?

Since the velocity of the hot-air balloon is given as (2.00 m/s)y, we can conclude that it is moving vertically upwards. The y-direction is usually considered to be the vertical direction, where positive values indicate upward motion and negative values indicate downward motion. Therefore, the balloon is moving upwards.

2) Determine the maximum height above the ground attained by the cork.

To find the maximum height attained by the cork, we can apply the principles of projectile motion. The horizontal velocity of the cork relative to the balloon is given as (6.00 m/s)x. Since there is no vertical velocity initially, we can assume the initial vertical velocity (Vy) of the cork is zero.

Using the equation for the vertical distance traveled by a projectile, we can determine the maximum height (H) reached by the cork:
H = (Vy^2) / (2g)

In this case, we plug in the values:
Vy = 0 (initial vertical velocity)
g = 9.8 m/s^2 (acceleration due to gravity)

H = (0^2) / (2 * 9.8)
H = 0 / 19.6
H = 0 meters

Therefore, the cork does not attain any height above the ground. It remains at a height of 7.00 meters above the ground, as it was when the cork was opened.

3) How long does the cork remain in the air?

To find the time of flight of the cork, we can use the equation for the time taken for a projectile to reach the ground:
t = 2 * Vy / g

In this case, Vy = 0 (initial vertical velocity) and g = 9.8 m/s^2 (acceleration due to gravity).

t = 2 * 0 / 9.8
t = 0 seconds

Therefore, the cork remains in the air for 0 seconds, or in other words, it immediately falls back to the ground after being expelled from the bottle.

Take up as positive Vf^2=vi^2+2gy

vf=(2,5)^2+2(-9,8)(-3)
vf=8,07m/s
vf=vi+gt
=(2,5)+(-9,8)t
t=0,57s