# Projectile motion beer slide problem

1. Oct 19, 2008

### haydn

1. The problem statement, all variables and given/known data

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 0.60 m from the base of the counter. The height of the counter is 0.900 m.

(a) With what velocity did the mug leave the counter?

(b) What was the direction of the mug's velocity just before it hit the floor?

2. Relevant equations

h=(vi^2*sin^2(x))/(2g)

r=(vi^2*sin(2x))/(g)

the variable x is the angle of the projectile

vi is the initial velocity

3. The attempt at a solution

I thought the angle of the projectile would be 0 since the mug is going straight off the bar but if I plug 0 in to the equation I get an answer that doesn't make sense. I also tried -45 but I got an incorrect answer. I think I'll be ok finding the answer if I can figure out the angle...

Thanks

2. Oct 20, 2008

### Philosophaie

I believe there is no initial velocity in the y-direction so only gravity to deal with. In the x-direction you have initial velocity and no gravity to deal with.

y=y0 - 1/2*g*t^2

y=0 when it hits the floor and y0 is the height of the counter, correct.

x=vi*t

vx=vi
vy=-1/2*g*t

Angle=atan(vy/vx)

3. Oct 20, 2008

### Redbelly98

Staff Emeritus
x is the unknown angle of the mug's trajectory as it hits the floor.

There are 2 unknown quantities in these 2 equations, so it can be solved with some algebraic manipulation and knowledge of trig formulas.

An alternative, and equally valid, approach would be to figure out how long it takes the mug to hit the floor.

4. Oct 21, 2008

### haydn

Alright. I'm having trouble with the algebraic manipulation and I don't think I got the correct answer. Could you solve it and tell me what you get?

Thanks a lot.

5. Oct 21, 2008

### Redbelly98

Staff Emeritus
Uh, no, it doesn't work that way here. Sorry!

If you solve each expression for vi^2, you could equate them and then try to find the angle.