Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 55: 117

Answer

Height: $5.5~m$ Initial speed: $11~m/s$

Work Step by Step

We know that $T=0.75s+\frac{0.75s}{2}$ $\implies T=1.125s$ The initial speed is given as $v_{f1}=v_{1i}-gT$ This can be rearranged as: as $v_{f1}=0m/s$ $\implies v_{i1}=gT$ We plug in the known values to obtain: $v_{i1}=(9.8m/s^2)(1.125s)$ $v_{i1}=11m/s$ Similarly $v_p=11m/s-(9.8m/s^2(0.75s))$ $v_p=3.7m/s$ Now $h=(\frac{v_{i1}+v_p}{2})t_1$ We plug in the known values to obtain: $h=(\frac{11+3.7}{2})(0.75s)$ $h=5.5m$
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.