You are in a hot air balloon (yes, another balloon problem!) rising from the gro

Question

You are in a hot air balloon (yes, another balloon problem!) rising from the ground at a constant velocity of 2.10 m/s upward. To celebrate the takeoff, you open a bottle of champagne, expelling the cork with a horizontal velocity of 6.40 m/s relative to the balloon. When opened, the bottle is 7.10 m above the ground What is the initial speed of the cork, as seen by your friend on the ground? Semit Ansuet Tries 0/e t the cork as seen by your What is the initial direction of the cork as seen by your friend? Give your answer as an angle relative to the horizontal Submit Answer Tries 0/8 Determine the maximum height of the cork above the ground. Submit AnswerTries 0/8 How long does the cork remain in the air? Submit AnswerTries 0/8

Explanation / Answer

The initial speed, v of the cork, as seen by your friend on the ground is given by

v = sqrt[2.1^2 + 6.4^2]

v = 6.45 m/s

the initial direction of the cork as seen by your friend as an angle, 'theta' with horizontal is given by

theta = tan^-1[2.1/6.4] = 18.2 degree

Maximum height of the cork above the ground = (7.10 + h ) m

v^2 = u^2 + 2ah

0 = u^2 + 2ah

h = -u^2 / 2a

h = -2.1^2 / (2 x -9.81)

h = 0.23 m

required height = 7.10 + 0.23 = 7.33 m

the cork remain in the air = 0.23/(average velocity of ascent) + sqrt[(2*7.33)/9.81]

= (0.23/1) + 1.22 = 1.45 s

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