# Finicky bartenders and the trajectory of a drinking glass.

1. Aug 19, 2007

### niyati

In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor 1.40 m from the base of the counter. If the height of the counter is .860 m, (a) with what velocity did the mug leave the counter, and (b) what was the direction of the mug's velocity just before it hit the floor?

(a) I have an equation, h = (vi^2(sinθi)^2)/2g, but I think it's wiser to use the equation of the vertical component of rf = ri + vit + .5a(t^2), which is yf = yi + vyit + .5a(t^2). The problem deals with the portion of the trajectory that has a negative velocity. For this portion, the final y coordinate is 0, the initial y coordinate is .860, and the initial velocity of y is 0, because it "starts" at the maximum height. Gravity is the driving force, and that is -9.8. So:

yf = yi + vyit + .5a(t^2)
0 = .860 + (0 * t) + .5(9.8)(t^2)
t = .41893938 seconds

This is the time is takes for the glass to basically complete the latter half of an imaginary parabola. Now, I don't know if I can assume this part. R is the range, or the total x component of a parabola. If this trajectory where to be a parabola, R would equal twice of 1.40, which is 2.80. And the total time of would ideally be twice the time it took to get from the highest point to the ground (2* .419) and that is .838. Given R = vxi(total time), the initial x component of velocity (which, I was told is constant in the chapter I am covering) is 3.34 (roundabout).

I think I'm stretching it a bit too much, and there is probably an easier way to doing that problem, but, well, that's what I thought of. If anyone would please tell me if my train of thought is correct, or, if it is incorrect, where did I go wrong? Can it be done this way? Why? Why not?

(b) I'm not sure how to figure this part out. Nor do I exactly know what this question is asking for.

2. Aug 19, 2007

### Mindscrape

Are you sure that the trajectory is going to be an entire parabola? The range formula doesn't apply to this case.

(a) In your y-distance equation, you found the time the mug is in the air for. Knowing that time, and how far the mug traveled in the x direction, can you figure out the velocity (which doesn't change in the x direction because there is no acceleration in the x direction)?

(b) If you know how fast the mug was going in the x direction when it hit the floor, and how fast it was going in the y direction when it hit the floor, how do you think you would be able to add the two directions (vectors) together?

3. Aug 20, 2007

### niyati

ugh. I knew I over-complicated it. (enough to where I forgot exactly WHAT velocity was...displacement over time)

So, the velocity when it shoots off the counter is equal to it's horizontal component, right? Because it is level and at its max height? In that case, vi = 3.34, which is exactly what I got before, but this method is...well, simpler. -_-''

As for (b), would this be all right?

vxf = vxi + at = 3.34 - 9.8(.419) = -.764i
vyf = vyi + at = 0 - 9.8(.419) = -4.1j

I don't think that's right, though. :S

4. Aug 20, 2007

### Mindscrape

Almost right. Are you sure there is an acceleration in the x direction? Also, I would assume that the problem wants the direction in the resultant form with an angle rather than the component form.

5. Aug 20, 2007

### niyati

>_<

ugh. I forgot to consider that. (bah, horizontal velocity is constant, and I still went ahead with the equation)

That makes a lot more sense, since the answer should be in the fourth quadrant.

[3.34, -4.1]

Or, 5.29 (magnitude) and about -51 degrees.

*crosses fingers*

Thank you so much for you help (and patience). :D

6. Aug 20, 2007

### niyati

your*

7. Aug 20, 2007

### Mindscrape

Yep, the speed is 5.29 m/s and the angle is 51 degrees below the horizontal. Also, even though it wasn't the most direct way, it was good that you found an alternate way for finding the initial x-velocity. Physics is full of alternative methods, and your method simply added factors of two to both sides of the equation, which makes for a good sanity check that two different valid views yield the same results.