## 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: 2

#### Answer

The rocket should be launched when the horizontal distance is 7.62 meters.

#### Work Step by Step

We can find the vertical acceleration of the rocket as; $\sum F = ma$ $F_{thrust}-mg=ma$ $a = \frac{F_{thrust}-mg}{m}$ $a = \frac{8.0~N-(0.500~kg)(9.80~m/s^2)}{0.500~kg}$ $a = 6.2~m/s^2$ We can find the time it takes for the rocket to reach a height of 20 meters; $y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2y}{a}}$ $t = \sqrt{\frac{(2)(20~m)}{6.2~m/s^2}}$ $t = 2.54~s$ We can find the horizontal distance the rocket travels in this time. Note that the rocket's horizontal speed will be equal to the cart's speed. $x = v_x~t$ $x = (3.0~m/s)(2.54~s)$ $x = 7.62~m$ The rocket should be launched when the horizontal distance is 7.62 meters.

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