A hot air balloon rises from the ground with a velocity of (2.0 m/s)yhat. A cham

Question

A hot air balloon rises from the ground with a velocity of (2.0 m/s)yhat. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (4.4 m/s)xhat relative to the balloon. When opened, the bottle is 6.2 m above the ground. (Neglect air resistance.)

(a) What is the speed of the cork, and its initial direction of motion, as seen by the same observer?

(b) Determine the maximum height above the ground attained by the cork.

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

Explanation / Answer

a) Cork has velocity of 2^2 + 4.4^2 = x^2 x = 4.8 m/s tan x = 2/4.4 angle= 24.4 above horizontal b) just use vertical velocity v^2 = vi^2 + 2ax 0 = (2)^2 + 2(-9.8)x x = .204 Add to initial height, 6.2 max height = 6.4 meters c) use kinematics x=xi + vt + .5at^2 again use only vertical 0 = 6.2 + 2t + (.5)(-9.8)t^2 t = 1.347 seconds

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