Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 3 - Kinematics in Two Dimensions - Problems - Page 74: 30

Answer

45.5 m/s

Work Step by Step

Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in vertical direction to the projectile. $\uparrow S=ut+\frac{1}{2}at^{2}$ Let's plug known values into this equation. $0=Vsin30^{\circ}t+\frac{1}{2}\times (-9.8\space m/s^{2})t^{2}$ $t=\frac{V}{9.8}s-(1)$ Let's apply the equation $S=ut$ in the horizontal direction to the projectile. $\rightarrow S=ut$ ; Let's plug known values into this equation. $183\space m=Vcos30^{\circ}t-(2)$ (1)=>(2), $183\space =\sqrt 3\frac{V}{2}\times\frac{V}{9.8}\space $ $V=\sqrt {\frac{183\times9.8\times2}{\sqrt 3}}m/s=45.5\space m/s$ Initial velocity of the projectile = 45.5 m/s
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.