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 2.60 m/s upward. To celebrate the takeoff, you open a bottle of champagne, expelling the cork with a horizontal velocity of 6.90 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?

What is the initial direction of the cork as seen by your friend? Give your answer as an angle relative to the horizontal.

Determine the maximum height of the cork above the ground.

How long does the cork remain in the air?

Explanation / Answer

velocity of the balloon relative to ground = Vbg = 2.6j

velocity of cork relative to balloon = Vcb = 6.9i

velocity of cork relative to ground = Vcg = Vcb + Vbg

Vcg = 6.9i + 2.6j

speed of the cork relative to your friend = sqrt(6.9^2+2.6^2) = 7.37 m/s   <<----answer

direction = tan^-1(2.6/6.9) = 20.6 degrees <<----answer

at maximum point final vertical speed of cork relative to ground

acceleration = a = -g

vf = 0

vi = 2.6

from equations of motion

vf^2 - vi^2 = 2*a*dy

dy = 2.6^2/(2*9.8) = 0.345

ymax - 6.9 = 0.345

ymax = 7.245 m <<----answer

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for the total journey the displacement y = -6.9m

vi = 2.6

a = -g

y = vi*t + 0.5*g*t^2

-6.9 = 2.6*t-0.5*9.8*t^2

t = 1.48 s <<----answer

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