## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 198: 1

#### Answer

The rocket should be launched when the horizontal distance is 38.9 meters.

#### Work Step by Step

We can find the acceleration of the rocket as; $\sum F = ma$ $F_{thrust}-mg=ma$ $a = \frac{F_{thrust}-mg}{m}$ $a = \frac{15.0~N-(0.800~kg)(9.80~m/s^2)}{0.800~kg}$ $a = 8.95~m/s^2$ We can find the time it takes for the rocket to reach a height of 30 meters; $y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2y}{a}}$ $t = \sqrt{\frac{(2)(30~m)}{8.95~m/s^2}}$ $t = 2.59~s$ We can find the distance the target travels in this time; $x = v~t$ $x = (15~m/s)(2.59~s)$ $x = 38.9~m$ The rocket should be launched when the horizontal distance is 38.9 meters.

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