anonymous
  • anonymous
PROJECTILE MOTION A bartender slides a beer mug at 1.50 m/s towards a customer at the end of a frictionless bar that is 1.20 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. a) How far away from the end of the bar does the mug hit the floor? b) What are the speed and direction of the mug at impact?
Physics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
I've answered Part A, and I got 0.75 m.
anonymous
  • anonymous
I'll show the work in a minute or two.
Michele_Laino
  • Michele_Laino
the velocity has two components: 1) the horizontal component whose magnitude is 1.50 m/s 2) the vertical component whose magnitude is: \[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = ...?\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I first found \(t\). \[\Delta y=V_0y t+ \frac{1}{2}at^2\]\[-1.2= -4.9t^2\]\(t=0.5s\) Then I found for \(\Delta x\)/. \[\Delta x=V_0xt\]\[\Delta x=(1.50)(0.5)\]\[\Delta x=0.75m\]
Michele_Laino
  • Michele_Laino
|dw:1444181078166:dw|
anonymous
  • anonymous
Wait, so should I find the y component first using the formula you showed or that's for the velocity?
Michele_Laino
  • Michele_Laino
since your mug, at impact, gains a vertical speed
anonymous
  • anonymous
And I got 5.077 m/s.
anonymous
  • anonymous
or 5.08 m/s
Michele_Laino
  • Michele_Laino
I got a different result: \[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = 4.85\]
Michele_Laino
  • Michele_Laino
do you agree?
anonymous
  • anonymous
No, because that's just the y component, and what I did it I found the y component (4.85) and the x component (1.5), then I used \(v=\sqrt {(4.85)^2+(1.5)^2}\).
Michele_Laino
  • Michele_Laino
ok! your answer is correct, I meant the vertical speed is 4.85
anonymous
  • anonymous
Ohh I was thinking ahead. Then to find the direction, should l just use tan^-1 V_fy/V_fx?
anonymous
  • anonymous
which gives me 72.81 degrees.
Michele_Laino
  • Michele_Laino
correct! for direction we can write this: \[4.85 = 1.5 \cdot \tan \theta \] |dw:1444181647493:dw|
anonymous
  • anonymous
Hold on, it would be negative, right?
Michele_Laino
  • Michele_Laino
no, since we consider this angle \(\beta=180-72.81=...\) |dw:1444181805577:dw|
Michele_Laino
  • Michele_Laino
sorry, you are right, it is -72.81 degree
Michele_Laino
  • Michele_Laino
|dw:1444181937222:dw|
anonymous
  • anonymous
Oh okay! It kinda confuses me when I don't see an illustration, but I just thought that since it's dropping, it would be in the IV Quadrant or something.
anonymous
  • anonymous
Thank you very much for helping me! This means a lot.I hope I'll ace the test tomorrow. :D
Michele_Laino
  • Michele_Laino
:) :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.