Question

You are in a hot air balloon (yes, another balloon problem!) rising from 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 4.40 m/s relative to the balloon. When opened, the bottle is 6.40 m above the ground.

(A)What is the initial speed of the cork, as seen by your friend on the ground?

(B)What is the initial direction of the cork as seen by your friend?

(C)Give your answer as an angle relative to the horizontal.

(D)Determine the maximum height of the cork above the ground.

(E)How long does the cork remain in the air?

Answer 1

A] speed = sqrt(4.4^2 + 2.2^2) = 4.92 m/s

B] initial direction = arctan(2.2/4.4) = 26.6 degree with horizontal

C] 26.6 degree with horizontal

D] maximum height = h+ v^2/2g = 6.4 + 2.2^2/[2*9.8] = 6.647 m

E] t is given by second equation of motion,

-6.4 = 2.2t - 0.5*9.8*t^2

t = 1.389 s answer