Projectile motion beer slide problem

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Homework Help Overview

The problem involves a beer mug sliding off a counter and striking the floor, with given distances and heights. It is situated in the context of projectile motion, specifically focusing on the initial velocity and trajectory angle of the mug as it falls.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial conditions of the mug's motion, questioning the angle of projection and the role of gravity in the vertical and horizontal components. Some suggest using kinematic equations to relate the height and distance, while others express uncertainty about algebraic manipulation and solving for unknowns.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to determine the initial velocity and angle. Some guidance has been offered regarding equating expressions for initial velocity, but there is no consensus on the method or solution yet.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to determine two unknowns (initial velocity and angle) from the provided equations and conditions. There is an emphasis on adhering to the forum's guidelines against providing direct solutions.

haydn
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Homework Statement



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?

Homework 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

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 into 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
 
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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)
 
haydn said:
h=(vi^2*sin^2(x))/(2g)

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

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.
 
Redbelly98 said:
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.

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.
 
haydn said:
Could you solve it and tell me what you get?

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.
 

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