Non Symetrical Parabola Question

[Husker70]

Homework Statement

In a local bar , a customer slides an empty beer mug down the counter for a refill. The bartender is just deciding to go home and rethink his life. He does not see the mug, which slides off the counter and strikes the floor 1.40m from the base of the counter. If the height of the counter is .860m. (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?

Homework Equations

(a) I got the correct answer for this by using the equation
yf= yi + vyit-1/2gt2
I got 3.3 m/s
(b) I would assume that since I know the height of the bar at .860m and the distance
traveled of 1.4m

The Attempt at a Solution

(b) The angle would make sense to me to be the tangent of opposite over adjcent but
I get 31.6 degrees, but from the book the answer is -50.9
Not sure what I'm doing wrong.

Thanks,
Kevin

 

[LowlyPion]

Homework Statement

In a local bar , a customer slides an empty beer mug down the counter for a refill. The bartender is just deciding to go home and rethink his life. He does not see the mug, which slides off the counter and strikes the floor 1.40m from the base of the counter. If the height of the counter is .860m. (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?

Homework Equations

(a) I got the correct answer for this by using the equation
yf= yi + vyit-1/2gt2
I got 3.3 m/s
(b) I would assume that since I know the height of the bar at .860m and the distance
traveled of 1.4m

The Attempt at a Solution

(b) The angle would make sense to me to be the tangent of opposite over adjcent but
I get 31.6 degrees, but from the book the answer is -50.9
Not sure what I'm doing wrong.

Let's see what is the title of this problem?
Non Symetrical Parabola Question

Does that suggest anything about what path the mug takes from bar top to floor? A triangle? Think more expansively.

What vertical velocity does it achieve? And how long to get there after leaving the bar top?

v_y²/(2*a) = Height
Height = 1/at²

Now you have the vertical velocity and you can figure the horizontal velocity from V_x = x/t those are the two components when it hits. (Of course V_x is the speed it leaves the counter.)

The Root Sum of the Squares then of the two components is the total magnitude and the two components determine direction (angle θ).

 

[Husker70]

That's beginning to make more sense.
I used VyF=Vyi-gt
So I get Vyf = 3.3m/s-(9.8m/s2).42s
Vyf = -.82m/s

Time I got by using Yf=Yi+Vyi(t)-1/2gt2
I then solved for t and get .42s

Not sure where to go from here.
Kevin

 

[Husker70]

Scratch the last. I didn't do that right

 

[Husker70]

I get Vyf to be -7.42 m/s
and Vyi to be 3.3 m/s
Are these right?

Kevin

 

[LowlyPion]

I get Vyf to be -7.42 m/s
and Vyi to be 3.3 m/s
Are these right?

Kevin

Answering these questions solves your problem.

How much time for the mug to drop to the floor?
Height = 1/2at²
.86 = 1/2 (9.8) t²
t = .419 sec

X Distance= 1.4 m
Horizontal velocity = 1.4/.419 = 3.34 m/s

Vertical velocity:
V² = 2*a*x
V² = 2*9.8*.86 = 4.11 m/s

Angle θ with the floor:

tan^-1(4.11/3.34) = Angle θ

 

[Husker70]

Thank you for the help.
I used the equation Vyf = Vyi -gt and got -7.42 m/s
Your equation makes a lot of sense but I couldn't find that one.
I'm not sure how to tell when to use which one. Is mine for a symetrical
parabola.
Is there an easy way to find the y value on a symetrical parabola. I have a total
diatance, the disatance at which I need the height, the initial velocity, and the initial
angle?
Thanks again,
Kevin

 

[LowlyPion]

Thank you for the help.
I used the equation Vyf = Vyi -gt and got -7.42 m/s
Your equation makes a lot of sense but I couldn't find that one.
I'm not sure how to tell when to use which one. Is mine for a symetrical
parabola.
Is there an easy way to find the y value on a symetrical parabola. I have a total
diatance, the disatance at which I need the height, the initial velocity, and the initial
angle?
Thanks again,
Kevin

Basically there are a couple of useful relationships, and it is likely more useful to understand these, because they allow you to partition a problem into pieces and combine what you need to solve for.

V = a* t
Acceleration times time is velocity.

V = x/t
Definition of velocity

x= 1/2 at²
the distance a body at rest falls related to time

V² = 2ax
relates the max velocity from rest under constant acceleration.

Now using those you can add initial conditions and develop more complex models based on what the question asks. But those are always a good place to start expanding the things you know about a problem as you move toward a solution.

Taking the more general form of these equations requires you to be careful about which time and which velocity is initial and final and can sometimes create more confusion than using the simpler forms and adding the terms that are relevant.