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 3.00 m/s upward. To celebrate the takeoff, you open a bottle of champagne, expelling the cork with a horizontal velocity of 6.70 m/s relative to the balloon. When opened, the bottle is 6.50 m above the ground.

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

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

3)Determine the maximum height of the cork above the ground.

4How long does the cork remain in the air?

Explanation / Answer

1)Use pythagorean theorem.
Vg=SQRT(Vb^2+Vc^2)
=SQRT(3^2+6.7^2)
=7.34m/s

2)Use trig ratios to solve this. The hypotenuse was the speed relative to the ground (6.5m/s). The side opposite of the angle above the horizontal is the orginal vertical speed (2.5m/s).
sin x=(opp/hyp)
=3/7.34
x= sin-1(3/7.34)
=24.12 degrees

3) For this question, the horizontal speed has no effect on the max height reached... only the vertical speed. given that the acceleration of gravity is 9.8 m/s^2:
V2^2=V1^2+2ad

we know that at the highest height reached, the speed will be 0.
0=V1^2+2ad
d=-V1^2/2a
=-(3)^2/2(-9.8)
=0.46m
then we have to add this to the height it was when it was released
h=6.5+0.46
=6.96m

4) d=v*t+a*t^2/2
-6.5=(3)t-4.9t^2
0=-4.9t^2+3t+6.5
now we use quadratic formula.

x=-(3)+/-SQRT(3^2-4(-4.9)(6.5)) /2(-4.9)
=13.61 or -14.22

Since x represents time, the negative root is extraneous and a purely mathematical construct. So it stays in the air for 13.61seconds.

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