A hot-air balloon is descending with a velocity of (2.00m/s)y^. A champagne bott

Question

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 5.55 m above the ground

Part 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.

Part B

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

Part C

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

Part D

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

Part E

How long does the cork remain in the air?

Explanation / Answer

a) Vx = 5.00 m/s; Vy = -2.00 m/s
b) V = [Vx²+Vy²] = 25+4 = 29 = 5.385 m/s
c) = arctan[2/5] degrees below horizontal
d) Y = 5.55 m (If you had said 'ascending', this would not be a trivial answer)
e) 5.55 = 2t + 4.9t2   --> t = 0.88 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.