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anonymous

  • one year ago

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?

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  1. anonymous
    • one year ago
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    I've answered Part A, and I got 0.75 m.

  2. anonymous
    • one year ago
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    I'll show the work in a minute or two.

  3. Michele_Laino
    • one year ago
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    the velocity has two components: 1) the horizontal component whose magnitude is 1.50 m/s 2) the vertical component whose magnitude is: \[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = ...?\]

  4. anonymous
    • one year ago
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    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 \(\Delta x\)/. \[\Delta x=V_0xt\]\[\Delta x=(1.50)(0.5)\]\[\Delta x=0.75m\]

  5. Michele_Laino
    • one year ago
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    |dw:1444181078166:dw|

  6. anonymous
    • one year ago
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    Wait, so should I find the y component first using the formula you showed or that's for the velocity?

  7. Michele_Laino
    • one year ago
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    since your mug, at impact, gains a vertical speed

  8. anonymous
    • one year ago
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    And I got 5.077 m/s.

  9. anonymous
    • one year ago
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    or 5.08 m/s

  10. Michele_Laino
    • one year ago
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    I got a different result: \[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = 4.85\]

  11. Michele_Laino
    • one year ago
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    do you agree?

  12. anonymous
    • one year ago
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    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 \(v=\sqrt {(4.85)^2+(1.5)^2}\).

  13. Michele_Laino
    • one year ago
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    ok! your answer is correct, I meant the vertical speed is 4.85

  14. anonymous
    • one year ago
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    Ohh I was thinking ahead. Then to find the direction, should l just use tan^-1 V_fy/V_fx?

  15. anonymous
    • one year ago
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    which gives me 72.81 degrees.

  16. Michele_Laino
    • one year ago
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    correct! for direction we can write this: \[4.85 = 1.5 \cdot \tan \theta \] |dw:1444181647493:dw|

  17. anonymous
    • one year ago
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    Hold on, it would be negative, right?

  18. Michele_Laino
    • one year ago
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    no, since we consider this angle \(\beta=180-72.81=...\) |dw:1444181805577:dw|

  19. Michele_Laino
    • one year ago
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    sorry, you are right, it is -72.81 degree

  20. Michele_Laino
    • one year ago
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    |dw:1444181937222:dw|

  21. anonymous
    • one year ago
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    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.

  22. anonymous
    • one year ago
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    Thank you very much for helping me! This means a lot.I hope I'll ace the test tomorrow. :D

  23. Michele_Laino
    • one year ago
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    :) :)

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