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A hot-air balloonist, rising vertically with a constant

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

Solution for problem 44E Chapter 2

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 44E

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (?Fig. E2.44?). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch ay-t, vy-t, and y-t graphs for the motion.

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Solution 44E Consider the formulas for position and velocity 2 x = x0+ v t0+ 1/2 at v = v0+ at Where x = final position x = original position 0 v0= original velocity a = acceleration v = final velocity (a). x = x + v t + 1/2 at2 0 0 2 2 x = 40.0 m + 5.0 m/s(0.250 s) + (0.5)( 9.8 m/s )(0.250 s) x = 41.556 m. (b). v = v + at 0 v = 5.0 m/s + ( 9.8 m/s )(0.250 s) v = 2.55 m/s. 2 (c).x = x 0 v t0+ 1/2 at x = 40.0 m + 5.0 m/s(1.00 s) + (0.5)( 9.8 m/s )(1.00s ) 2 x = 49.9 m. (d).v = v 0 at v = 5.0 m/s + ( 9.8 m/s )(1.00 s) v = 4.8m/s. (e).x = 0 to solve for time. 2 0 = 40 + 5(t) + 1/2( 9.8)t Using quadratic formula. t = b ± 4ac/2a Here a = 9.8 a = 5 a = 40 t = 9.166 s Using this time put into velocity formula v = v + at 0 v = 5 + ( 9.8)(9.166) v = 84.83 s To find greatest height,we need to find the time it takes for velocity to reach 0 m/s. So we can take final velocity as zero. 0 = 5 + ( 9.8)t t = 0.510204 s Put this time into position formula x = x + v t + 1/2 at 2 0 0 x = 40.0 m + 5.0 m/s(0.510204 s) + (0.5)( 9.8 m/s )(0.510204 s) 2 x = 41.2775 m Time v/s velocity graph is shown below. Y v/s t graph is as shown below.

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Chapter 2, Problem 44E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

Since the solution to 44E from 2 chapter was answered, more than 1029 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: Ground, sandbag, velocity, release, strike. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 44E from chapter: 2 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. The answer to “A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (?Fig. E2.44?). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch ay-t, vy-t, and y-t graphs for the motion.” is broken down into a number of easy to follow steps, and 105 words. This textbook survival guide was created for the textbook: University Physics, edition: 13.

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A hot-air balloonist, rising vertically with a constant