ent FULL scREEN PRINTERVERSION BACK NE Chapter 02, Problem 044 Your answer is pa

Question

ent FULL scREEN PRINTERVERSION BACK NE Chapter 02, Problem 044 Your answer is partially correct. Try again. A startled armadillo leaps upward, rising 0.584 m in the first o.191 s (a) what is its initial speed as it leaves the ground? (b) what is its speed at the height of 0.584 m? (c) How much higher does it go? Use g 9.81 m/s (a) Numbe 3.994 Units T m/s (b) Number 12.123 UnitsT m/s (c) Number 1.268 Un Click if you would like to show Work for this question Open Show Work SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM VIDEO MINI LECTURE

Explanation / Answer

Problem 044:

Let ,

u = initial upward leap speed of armadillo

v = Final upward leap speed of armadillo after first 0.191 s

(a)Average speed (vavg) of armadillo in first 0.191 s= 0.584/0.191 = 3.06 m/s

i.e. vavg = (u+v)/2

u + v = 2× 3.06 = 6.12 m/s. ............. (1)

u - v = g(0.191) = 9.81×0.191 = 1.87 ................(2)

solving (1) & (2) , we have

u = 3.99 m/s

(b) speed at 0.584 m

vh= 6.12- 3.99

= 2.12 m/s

(c) time for armadillo to reach its max height ,

t = u/g = 3.99/9.81 = 0.40714 s

Max height for armadillo H = (1/2) gt2

H = (1/2).9.81.(0.40714)2 = 0.813 m

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