Question

4-14)A hot-air balloon is descending with a velocity of (−2.00m/s)y^. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (5.00m/s)x^ relative to the balloon. When opened, the bottle is 5.35 m above the ground. Please help! i dont understand how to do it. This was the only information given

a.What is the initial velocity of the cork, as seen by an observer on the ground? Give your answer in terms of the x and y unit vectors.

Express your answer using three significant figures.

b.

What is the speed of the cork as seen by the same observer?

Express your answer using three significant figures.

c. What is initial direction of motion as seen by the same observer?

Express your answer using three significant figures.

d.

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

Express your answer using three significant figures.

e.

How long does the cork remain in the air?

Express your answer using two significant figures.

Answer 1

The velocity of the cork can be treated with x- and y-components.
Since the cork is rising with the balloon the y-component for the
cork is also - 2.00 m/s. x-component is 5.00 m/s:
Vx = 5.00 m/s
Vy = - 2.00 m/s

A). V = 5 I - 2 j m/s

B). V = √(V_x²+V_y²) = √(5²+(-2)²) = 5.385 m/s

C). tan θ = 2.00/5.00 => θ = arctan( 0.4) = 21.8° (above -ve x-axis)

D). The x-direction has no acceleration, so the horizontal velocity stays constant.
Max height is when rock turns and the y-component becomes zero  
Vyf^2 = Vyi^2 - 2gd
d = (Vyf^2-Vyi^2) / (-2g) = (0² - (-2)²) / (-2 x 9.81) = 0.204 m
Since the bottle was 5.35 m above ground, max height is 5.554 m