Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 64: 19

Answer

total time during which ball remain in air $=t=3.28s $

Work Step by Step

When the ball will come back to its intial position that is 2m above from ground its velocity will be same but in opposite direction. Net displacement will be zero. let $t_1$ be the total time taken by the ball to reach to its intial height. We have $\vec s_1=0$ $\vec u=15m/s$ $\vec v=-15m/s$ $\vec a=-g =-9.8m/s^2$ We know that $\vec s_1 = \vec ut_1 +\frac{1}{2}\vec a (t_1)^2$s $0= 15t_1 +\frac{1}{2}(-9.8)(t_1)^2$s $30t_1 =9.8( t_1)^2$s $t_1 =\frac{30}{9.8} $s $t_1 =3.06$s let $t_2 $ be the time taken from 2m height to ground. here $\vec s_2 =2m \\ \vec a= 9.8 m/s^2 \\ \vec u= 15 m/s$ $\therefore \vec s_2 = \vec ut_2 + \frac{1}{2}\vec a(t_2)^2$ $ 2 = 15 t _2+ 4.9 (t_2)^2$ $t_2 = 0.22$s thus total time during which ball remain in air $=t=t_1+t_2 =3.06s + 0.22s =3.28s $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.