### Cork of a Champagne Bottle Pops out, Rises Vertically, and fall to the Floor

A champagne bottle is held upright 1.2 m above the floor as the wire around its cork is removed. The cork then pops out, rises vertically, and falls to the floor 1.4 s later.
What was the cork's initial velocity?
 a) 3.2 m/s b) 4.9 m/s c) 7.3 m/s N d) 6.0 m/s e) None of the above.
What height above the bottle did the cork reach?
 a) 2.12 m b) 1.43 m c) 2.21 m N d) 1.84 m e) None of the above.
Solution:

Solution:

For solving this problem, use can be made of the time needed by the cork to reach the floor, t = 1.4 s, which we know is 1.2 m below the launching point of the cork. This quantity correspond to the cork displacement of y = -1.2 mhe initial velocity of the cork can be calculated using the formula g is the acceleration of gravity. Thus, solving for the initial velocity, v0y, in the previous relation,  . And the initial velocity becomes .

Knowing the initial velocity of the cork, the maximum height can be calculated from the formula, , (F_F 5), since at the velocity of the cork is zero, . The maximum height can be calculated, . Substituting the numerical values .

its final velocity?
 a) - 6.5 m/s N b) - 7.7 m/s c) - 8.5 m/s d) - 5.4 m/s e) None of the above.
Solution:

At the floor level, the final velocity of the cork is obtained directly from the formula that evaluated at the known numerical values results in  .

The numerical calculation with MathCad is

 by Luis F. Sáez, Ph. D. Comments and Suggestions: LSaez@dallaswinwin.com