Chapter 02, Problem 044 A startled armadillo leaps upward, rising 0.504 m in the

Question

Chapter 02, Problem 044 A startled armadillo leaps upward, rising 0.504 m in the first 0.204 (a) What is its initial speed as it leaves the ground? (b) what is its speed at the height of 0.504 m? (c) How much higher does go? useg m/s2. (a) Number Units (b) Number Units Units (c) Number click if you would like to show work for this question: open show Work Question Attemptsa o of 6 used SAVE FOR LATER SURAEITANSIVER Copyright D 2000-2017 by John Wiley Sons, Inc. or related companies. All rights reserved. All Rights Reserved. A Devision MacBook Air

Explanation / Answer

a) we knows that S = ut +1/2at2

0.504=0.204 u - 0.5*9.8*0. 2042 ( a = the acceleration due to gravity which is -9.8 m/s^2 )

u = 3.47 m/s

b) we knows that v =u+at

= 3.47 -9.8*.204 = 1.47 m/s

c) For the final height, the v is zero,

therefore again using v=u- 9.8 t

or t = 3.47/9.8 = 0.354 s

putting the value of t in displacement equation

we get S = 3.47*0.354 - 0.5*9.8*0.3542

=0.614m

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