A hot-air ballon rises from the ground with a velocity of (2.00 m/s)y. A champag

Question

A hot-air ballon 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 (5.00 m/s)x, relative to the ballon. When opened, the bottle is 6.00 m above the ground. (a) What is the initial velocity of the cork, as seen by an observer on the ground? (b) What is the speed of the cork, and its initial direction of motion, as seen by the same observer? (c) Determine the maximum height above the ground attained by the cork. (d) How long does the cork remain in air?

Explanation / Answer

Velocity of hot air ballon Vhb = (2.00m/s)y Velocity of cork Vc = (5.00m/s)x Initial height of the bottle Yi = 6.00 m a) The velocity of the cork is relative to ground is            Vc relative to g = Velocity of ballon + velocity of cork                                    = (2.00m/s)y + (5.00m/s)x      b) magnitude of the speed of the cork Vcg =(Vc2 + Vhb2)                                                                = (2.002 + 5.002)                                                                  = 5.38 m/s       Direction = tan-1 (Vc /Vhb)                       = tan-1 (5 / 2)                       = 68.19o c) maximum height Yf = Vcg 2/2g + 6.00m                                  = 5.38 * 5.38 / 2*9.8 + 6.00                                   = 7.48 m

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