Asked by Jan
A hot-air balloon rises from the ground with a velocity of (2.00 m/s)y. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (6.00 m/s)x relative to the balloon. When opened, the bottle is 7.00 m above the ground.
1)What is its initial direction of motion, as seen by the same observer?
2)Determine the maximum height above the ground attained by the cork
3)How long does the cork remain in the air?
1)What is its initial direction of motion, as seen by the same observer?
2)Determine the maximum height above the ground attained by the cork
3)How long does the cork remain in the air?
Answers
Answered by
Msibi ntokozo
Take up as positive Vf^2=vi^2+2gy
vf=(2,5)^2+2(-9,8)(-3)
vf=8,07m/s
vf=vi+gt
=(2,5)+(-9,8)t
t=0,57s
vf=(2,5)^2+2(-9,8)(-3)
vf=8,07m/s
vf=vi+gt
=(2,5)+(-9,8)t
t=0,57s
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