Answer Check for kinematic in 2-d

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The discussion revolves around a physics problem involving the motion of a beer mug sliding off a counter. The correct initial velocity of the mug as it leaves the counter is determined to be 2.68 m/s. For the direction of the mug's velocity just before impact, the initial calculations were incorrect, leading to confusion about the angle. The x-component of the velocity remains constant at 2.68 m/s, while the y-component is recalculated to be 4.01 m/s, affecting the impact angle. The participants emphasize the importance of correctly applying kinematic equations to resolve the discrepancies in the y-velocity.
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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 1.10 m from the base of the counter. The height of the counter is 0.820 m.
(a) With what velocity did the mug leave the counter?
correct check mark m/s
(b) What was the direction of the mug's velocity just before it hit the floor?
wrong check mark
Your answer differs from the correct answer by 10% to 100%.° (below the horizontal)

Homework Equations


I found (a) which is 2.68 m/s


The Attempt at a Solution



For b) i used sqrt(2.68^2+2.68^) = 3.79 and did,
theta = cos^-1(2.68/3.79) which is 45 degrees, which is wrong but since it say below the horizontal axis, will it be -315 degree?
 
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Filenes said:

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 1.10 m from the base of the counter. The height of the counter is 0.820 m.
(a) With what velocity did the mug leave the counter?
correct check mark m/s
(b) What was the direction of the mug's velocity just before it hit the floor?
wrong check mark
Your answer differs from the correct answer by 10% to 100%.° (below the horizontal)

Homework Equations


I found (a) which is 2.68 m/s


The Attempt at a Solution



For b) i used sqrt(2.68^2+2.68^) = 3.79 and did,
theta = cos^-1(2.68/3.79) which is 45 degrees, which is wrong but since it say below the horizontal axis, will it be -315 degree?

I got the same for part a but I got something different for the impact angle.

the impact creates a triangle with x velocity and y velocity at impact. x velocity is constant at 2.68 m/s and I get 4.01 m/s for y velocity. Just finish it and you should get the right answer.
 
thanks alot.. now i got to figure out why is the y velocity different .. thanks!
 

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