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A shot putter releases the shot some distance above the

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 18E Chapter 3

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 18E

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0o above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. (a) What are the components of the shot’s acceleration while in flight? (b) What are the components of the shot’s velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for R in Example 3.8 ?not? give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? (f) Draw x-t, y-t, vx-t, and vy-t graphs for the motion.

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Solution 18E (a) The horizontal acceleration is always zero as the horizontal motion of the shot is not influenced by gravity. 2 The vertical component of acceleration is 9.8 m/s . (b) The velocity of projection is v = 12.0 m/s At the beginning, The x-component of velocity v = 12.0 m/s × cos 51 0 xi v = 7.55 m/s xi 0 The y-component of velocity v yi= 12.0 m/s × sin 51 v yi= 9.32 m/s At the end, The x-component always remains the same as the gravity does not act on it. Therefore, v = 7.55 m/s xf The y-component can be calculated as follows. v = v gt yf yi 2 v yf= 9.32 m/s 9.8 m/s × 2.08 s (time = 2.08 s) v yf= 11.06 m/s (c) Horizontal component of velocity = 7.55 m/s Time = 2.08 s Horizontal distance travelled by the shot = 7.55 m/s × 2.08 s = 15.70 m Horizontal range can also be...

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Chapter 3, Problem 18E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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