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Tutorial Exercise In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.35 m. The mug slides off the counter and strikes the floor 0.90 m from the base of the counter. (a) with what velocity did the mug leave the counter? (b) What was the direction of the mug's velocity just before it hit the floor? Part 1 of 6 - Conceptualize Based on our everyday experiences and the description of the problem, a reasonable speed of the mug would be a few m/s. It will hit the floor at an angle between 0° and 90°, probably closer to 45°. Part 2 of 6 - Categorize We are looking for two different velocity components, but we are only given two distances. Our approach will be to separate the vertical and horizontal motion into components. By using the height the mug falls, we can find the time interval of the fall. Once we know the time, we can find the horizontal and vertical components of the velocity, and from those values, we can calculate the angle of impact. For convenience, we will set the origin to be the point where the mug leaves the counter. Part 3 of 6 - Analyze (a) The mug leaves the counter horizontally with a velocity denoted by Vxi. Since there is no horizontal acceleration, the time t at which it hits the ground is given by xf = Vxit. So we have = Vxit. Your response differs from the correct answer by more than 10%. Double check your calculations. m Solving for t gives us the following expression in terms of Vxi. t = Your response differs from the correct answer by more than 10%. Double check your calculations. m Vxi Since the mug falls with constant acceleration, we can use the following equation. y = yi + Vyit + ayt² The mug slides off the counter in the x direction, both initial y components for displacement and velocity are zero. Substituting the values we know into the equation above, gives the following. | m = 0 + 0.t - 1 (9.80 m/s²)² Now substitute the expression we found for t above. 0 = m- -글(1 Submit Skip (you cannot come back) 2 m/s² m Vxi

Tutorial Exercise In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.35 m. The mug slides off the counter and strikes the floor 0.90 m from the base of the counter. (a) with what velocity did the mug leave the counter? (b) What was the direction of the mug's velocity just before it hit the floor? Part 1 of 6 - Conceptualize Based on our everyday experiences and the description of the problem, a reasonable speed of the mug would be a few m/s. It will hit the floor at an angle between 0° and 90°, probably closer to 45°. Part 2 of 6 - Categorize We are looking for two different velocity components, but we are only given two distances. Our approach will be to separate the vertical and horizontal motion into components. By using the height the mug falls, we can find the time interval of the fall. Once we know the time, we can find the horizontal and vertical components of the velocity, and from those values, we can calculate the angle of impact. For convenience, we will set the origin to be the point where the mug leaves the counter. Part 3 of 6 - Analyze (a) The mug leaves the counter horizontally with a velocity denoted by Vxi. Since there is no horizontal acceleration, the time t at which it hits the ground is given by xf = Vxit. So we have = Vxit. Your response differs from the correct answer by more than 10%. Double check your calculations. m Solving for t gives us the following expression in terms of Vxi. t = Your response differs from the correct answer by more than 10%. Double check your calculations. m Vxi Since the mug falls with constant acceleration, we can use the following equation. y = yi + Vyit + ayt² The mug slides off the counter in the x direction, both initial y components for displacement and velocity are zero. Substituting the values we know into the equation above, gives the following. | m = 0 + 0.t - 1 (9.80 m/s²)² Now substitute the expression we found for t above. 0 = m- -글(1 Submit Skip (you cannot come back) 2 m/s² m Vxi

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Tutorial Exercise In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.35 m. The mug slides off the counter and strikes the floor 0.90 m from the base of the counter. (a) with what velocity did the mug leave the counter? (b) What was the direction of the mug's velocity just before it hit the floor? Part 1 of 6 - Conceptualize Based on our everyday experiences and the description of the problem, a reasonable speed of the mug would be a few m/s. It will hit the floor at an angle between 0° and 90°, probably closer to 45°. Part 2 of 6 - Categorize We are looking for two different velocity components, but we are only given two distances. Our approach will be to separate the vertical and horizontal motion into components. By using the height the mug falls, we can find the time interval of the fall. Once we know the time, we can find the horizontal and vertical components of the velocity, and from those values, we can calculate the angle of impact. For convenience, we will set the origin to be the point where the mug leaves the counter. Part 3 of 6 - Analyze (a) The mug leaves the counter horizontally with a velocity denoted by Vxi. Since there is no horizontal acceleration, the time t at which it hits the ground is given by xf = Vxit. So we have = Vxit. Your response differs from the correct answer by more than 10%. Double check your calculations. m Solving for t gives us the following expression in terms of Vxi. t = Your response differs from the correct answer by more than 10%. Double check your calculations. m Vxi Since the mug falls with constant acceleration, we can use the following equation. y = yi + Vyit + ayt² The mug slides off the counter in the x direction, both initial y components for displacement and velocity are zero. Substituting the values we know into the equation above, gives the following. | m = 0 + 0.t - 1 (9.80 m/s²)² Now substitute the expression we found for t above. 0 = m- -글(1 Submit Skip (you cannot come back) 2 m/s² m Vxi
Transcribed Image Text:Tutorial Exercise In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.35 m. The mug slides off the counter and strikes the floor 0.90 m from the base of the counter. (a) with what velocity did the mug leave the counter? (b) What was the direction of the mug's velocity just before it hit the floor? Part 1 of 6 - Conceptualize Based on our everyday experiences and the description of the problem, a reasonable speed of the mug would be a few m/s. It will hit the floor at an angle between 0° and 90°, probably closer to 45°. Part 2 of 6 - Categorize We are looking for two different velocity components, but we are only given two distances. Our approach will be to separate the vertical and horizontal motion into components. By using the height the mug falls, we can find the time interval of the fall. Once we know the time, we can find the horizontal and vertical components of the velocity, and from those values, we can calculate the angle of impact. For convenience, we will set the origin to be the point where the mug leaves the counter. Part 3 of 6 - Analyze (a) The mug leaves the counter horizontally with a velocity denoted by Vxi. Since there is no horizontal acceleration, the time t at which it hits the ground is given by xf = Vxit. So we have = Vxit. Your response differs from the correct answer by more than 10%. Double check your calculations. m Solving for t gives us the following expression in terms of Vxi. t = Your response differs from the correct answer by more than 10%. Double check your calculations. m Vxi Since the mug falls with constant acceleration, we can use the following equation. y = yi + Vyit + ayt² The mug slides off the counter in the x direction, both initial y components for displacement and velocity are zero. Substituting the values we know into the equation above, gives the following. | m = 0 + 0.t - 1 (9.80 m/s²)² Now substitute the expression we found for t above. 0 = m- -글(1 Submit Skip (you cannot come back) 2 m/s² m Vxi
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