Question

The cork from a champagne bottle slip through the hands of the waiter opening it, moving with an initial velocity v=11.8 m/s at an angle of 76.4 degrees above horizontal. A diner is sitting a horizontal distance d away when this happens. Assume the cork leaves the waiter's hands at the same vertical level as the diner and that the cork falls back to this vertical level when it reaches the diner
Randomized variables
v_o = 11.8 m/s
theta = 76.4 degrees
a) Calculate the time, t_d in seconds, for the cork to reach the diner
b) Reacting quickly to avoid being struck, the diner moves 2.00m

Answer 1

horizontally directly toward the waiter opening the champagne bottle. Determine the horizontal distance, d in meters, between the person and the diner at the time the cork reaches where the diner had previously been sitting.

Answer 2

Any help is appreciated!!!! :( I am very lost... My professor has not gone over this.

Answer 3

the situation can be represented as below:

along the horizontal or \(x-\) axis we have an uniform motion with constant velocity, so the requested time, is:
\[\Large {t_d} = \frac{d}{{{v_0}\cos \theta }}\]