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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've answered Part A, and I got 0.75 m.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll show the work in a minute or two.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the velocity has two components: 1) the horizontal component whose magnitude is 1.50 m/s 2) the vertical component whose magnitude is: \[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444181078166:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, so should I find the y component first using the formula you showed or that's for the velocity?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since your mug, at impact, gains a vertical speed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I got 5.077 m/s.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got a different result: \[\sqrt {2gh} = \sqrt {2 \cdot 9.81 \cdot 1.2} = 4.85\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, 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
 one year ago
Best ResponseYou've already chosen the best response.1ok! your answer is correct, I meant the vertical speed is 4.85

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh I was thinking ahead. Then to find the direction, should l just use tan^1 V_fy/V_fx?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which gives me 72.81 degrees.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! for direction we can write this: \[4.85 = 1.5 \cdot \tan \theta \] dw:1444181647493:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hold on, it would be negative, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since we consider this angle \(\beta=18072.81=...\) dw:1444181805577:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry, you are right, it is 72.81 degree

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444181937222:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much for helping me! This means a lot.I hope I'll ace the test tomorrow. :D
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