## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 43

#### Answer

a. 10.4m/s, 17$^o$ above the horizontal. b. 10.4m/s, 17$^o$ below the horizontal.

#### Work Step by Step

a. Call the upward direction positive. The velocity of the ball relative to the ground is the vector sum of the horizontal and vertical motions. From the given information, we see that the velocity relative to the ground is (10 m/s, 3.0 m/s). Find the magnitude using the Pythagorean Theorem and the direction using the definition of the tangent. $$v_{BG} = 10.4 m/s$$ The angle is at 17$^o$ above the horizontal. b. The only change is that the balloon is descending, so the velocity relative to the ground is (10 m/s, -3.0 m/s). Find the magnitude using the Pythagorean Theorem and the direction using the definition of the tangent. $$v_{BG} = 10.4 m/s.$$ The angle is now at 17$^o$ below the horizontal.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.