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(13%) Problem 3: While watching the clouds pass by, you notice a European swallow flying horizontally at a height of h = 22.65 m. When the swallow is directly overhead, it drops an m = 11.7 kg coconut. Refer to the diagram. From your ornithological studies, you know that the air- speed of this particular species of swallow, while carrying such a load, is vo=7.88 m/s. Neglect air resistance. Assume that your head is at the origin of the coordinate system. 20 ▲ 20% Part (a) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut, as observed by you, at the moment it is released directly overhead. =▲ 20% Part (b) Let be the time-dependent position vector of the coconut with time measured from the instant it was released by the swallow. Enter an expression for its horizontal component as a function of time. 20% Part (c) Enter an expression, in unit vector notation, for the time-dependent velocity vector, (t), of the coconut. 20% Part (d) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut as observed by you at t = 1.0 s. L=2544 kg-m²/s ✓Correct! 20% Part (e) Calculate the magnitude, in kilogram squared meters per second, of the coconut as observed by you immediately before it hits the ground. L= kg - m²/s sin cos tan π cotan asin() acos() E ( ) 7 8 9 M4 5 6 HOME atan acotan() sinh 1 2 3 cosh tanh cotanh + - 0 END Degrees Radians NO BACKSPACE DEL CLEAR

(13%) Problem 3: While watching the clouds pass by, you notice a European swallow flying horizontally at a height of h = 22.65 m. When the swallow is directly overhead, it drops an m = 11.7 kg coconut. Refer to the diagram. From your ornithological studies, you know that the air- speed of this particular species of swallow, while carrying such a load, is vo=7.88 m/s. Neglect air resistance. Assume that your head is at the origin of the coordinate system. 20 ▲ 20% Part (a) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut, as observed by you, at the moment it is released directly overhead. =▲ 20% Part (b) Let be the time-dependent position vector of the coconut with time measured from the instant it was released by the swallow. Enter an expression for its horizontal component as a function of time. 20% Part (c) Enter an expression, in unit vector notation, for the time-dependent velocity vector, (t), of the coconut. 20% Part (d) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut as observed by you at t = 1.0 s. L=2544 kg-m²/s ✓Correct! 20% Part (e) Calculate the magnitude, in kilogram squared meters per second, of the coconut as observed by you immediately before it hits the ground. L= kg - m²/s sin cos tan π cotan asin() acos() E ( ) 7 8 9 M4 5 6 HOME atan acotan() sinh 1 2 3 cosh tanh cotanh + - 0 END Degrees Radians NO BACKSPACE DEL CLEAR

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
Problem 42P: A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.70 m/s...
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(13%) Problem 3: While watching the clouds pass by, you notice a European swallow flying horizontally at a height of h = 22.65 m. When the swallow is directly overhead, it drops an m = 11.7 kg coconut. Refer to the diagram. From your ornithological studies, you know that the air- speed of this particular species of swallow, while carrying such a load, is vo=7.88 m/s. Neglect air resistance. Assume that your head is at the origin of the coordinate system. 20 ▲ 20% Part (a) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut, as observed by you, at the moment it is released directly overhead. =▲ 20% Part (b) Let be the time-dependent position vector of the coconut with time measured from the instant it was released by the swallow. Enter an expression for its horizontal component as a function of time. 20% Part (c) Enter an expression, in unit vector notation, for the time-dependent velocity vector, (t), of the coconut. 20% Part (d) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut as observed by you at t = 1.0 s. L=2544 kg-m²/s ✓Correct! 20% Part (e) Calculate the magnitude, in kilogram squared meters per second, of the coconut as observed by you immediately before it hits the ground. L= kg - m²/s sin cos tan π cotan asin() acos() E ( ) 7 8 9 M4 5 6 HOME atan acotan() sinh 1 2 3 cosh tanh cotanh + - 0 END Degrees Radians NO BACKSPACE DEL CLEAR
Transcribed Image Text:(13%) Problem 3: While watching the clouds pass by, you notice a European swallow flying horizontally at a height of h = 22.65 m. When the swallow is directly overhead, it drops an m = 11.7 kg coconut. Refer to the diagram. From your ornithological studies, you know that the air- speed of this particular species of swallow, while carrying such a load, is vo=7.88 m/s. Neglect air resistance. Assume that your head is at the origin of the coordinate system. 20 ▲ 20% Part (a) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut, as observed by you, at the moment it is released directly overhead. =▲ 20% Part (b) Let be the time-dependent position vector of the coconut with time measured from the instant it was released by the swallow. Enter an expression for its horizontal component as a function of time. 20% Part (c) Enter an expression, in unit vector notation, for the time-dependent velocity vector, (t), of the coconut. 20% Part (d) Calculate the magnitude, in kilogram squared meters per second, of the angular momentum of the coconut as observed by you at t = 1.0 s. L=2544 kg-m²/s ✓Correct! 20% Part (e) Calculate the magnitude, in kilogram squared meters per second, of the coconut as observed by you immediately before it hits the ground. L= kg - m²/s sin cos tan π cotan asin() acos() E ( ) 7 8 9 M4 5 6 HOME atan acotan() sinh 1 2 3 cosh tanh cotanh + - 0 END Degrees Radians NO BACKSPACE DEL CLEAR
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