You are in a not air balloon (yes, another balloon problem!) rising om the groun

Question

You are in a not air balloon (yes, another balloon problem!) rising om the ground at a constant velocity of 2.20 m/s upward. To celebrate the takeoff, you open a bottle of champagne, expelling the cork with a horizontal velocity of 4.90 m/s relative to the balloon. When opened, the bottle is 6.90 m above the ground. What is the initial speed of the cork, as seen by your friend on the ground? Submit Answer Tries 0/8 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 Answer Tries 0/8 How long does the cork remain in the air? Submit Answer Tries 0/8

Explanation / Answer

Speed of Balloon = 2.2 m/s upward
Speed of Cork , relative to baloon = 4.90 m/s Horizontal

Initial speed of cork, as seen ny friend = sqrt ((Speed of Cork ,relative to baloon)^2 + (Speed of Balloon)^2)
Initial speed of cork, as seen ny friend = sqrt(4.9^2 + 2.2^2) m/s
Initial speed of cork, as seen ny friend = 5.37 m/s

Tan (theta) = 2.2/4.9
theta = tan^-1(2.2/4.9)
Initial direction of the cork = 24.18o

Vertical Velocity = 2.2m/s
At Max Height Vf = 0
vf^2 = u^2 - 2 * g * h
h = 2.2^2 / 2*9.8
h = 0.247 m

Maximum Height of cork above the ground = 6.9 + 0.247 = 7.147 m

time taken to reach this height t1 = 2.2/9.8 s = 0.224 s

Now, Cork starts coming down -
Total time taken by cork to come down from height h = 7.147m =
v^2 = 0 + 2 * g *h
v = sqrt(2 * 9.8 * 7.147) m/s
v = 11.84 m/s

t2 = v/g
t2 = 11.84/9.8 s
t2 = 1.21

Total time cork is in the air = t1 + t2 = 0.224 s + 1.21 s
Total time cork is in the air = 1.434 s

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