# Cork flying into air, only acceleration and total time

Tags:
1. Jun 17, 2016

### eric_cravin

1. The problem statement, all variables and given/known data
Bindi pops the cork off a bottle of champagne. The cork travels vertically into the air. She notices that the cork takes 4s to return to starting position.
a) how long does cork take to reach maximum height?

2. Relevant equations
I know that a= -9.81m/s/s, and the final velocity would be is 0m/s, but I'm not sure how to work out solution or equation to find solution.

3. The attempt at a solution
I drew a picture but I still have no clue.

2. Jun 18, 2016

### late347

I was wondering if

Bindi has a stopwatch.
does 4 seconds include the upwards and downwards travelling of the cork?

or does 4 seconds only include downward falling from apex to ground?

3. Jun 18, 2016

### billy_joule

That's not right. The cork Is falling and so had some velocity.

I'd say it's the former.
Also note that the starting position isn't the ground (unless the bottle is subterranean..).
Throw your pen up a few times, does the upward part of the flight take more, less or equal time as the downward part? What SUVAT equation can you use to find out for sure? You know total displacement is zero when the pen (or cork) returns to the starting position.

4. Jun 18, 2016

### late347

cork has some launch velocity $$v_0= ~~something ~~positive$$

cork displaces zero meters.

What goes up comes down. Cork displaces zero. First the cork goes up distance x. Then the cork returns distance x to earth.

We are assuming level floor and no air resistance. Launch angle will be 90deg from the horizontal floor. Time taken for this roundabout journey in the sky equals 4 seconds overrall time. It took me some time to realize this factoid but it seems to be legitimate interpretation.

real champagne bottle corks probably don't stay in the air for 4 seconds total time.

launch velocity of the cork can be calculated from

$$displacement~~=v_o*t~~+0.5*a*t^2$$

time for deceleratoin can be calculated from

$v_1= v_0+at$

where v1= 0 m/s
a= -9.81
v0 = ???

5. Jun 18, 2016

### CWatters

You don't need any of that to solve this problem. Try thinking about the symmetry of the problem. The acceleration has same magnitude on the way up and down. The displacement is the same. The final velocity on the way up is the initial velocity on the way down. The initial velocity on the way up is the same as the final velocity on the way down. With all this symmetry what do you think the time for each phase is likely to be?