(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

When startled, an armadillo will leap upward. Suppose it rises 0.544 m in the first .2 s. (a) What is its initial speed as it leaves the ground? (b) what is its speed at the height of .544 m? (c) How much higher does it go?

2. Relevant equations

/\X = Vot+.5at^2

Vf^2=Vo^2+2a/\x

3. The attempt at a solution

(a) /\x = .544 m at .2 seconds so

.544 = Vo (0.2) + .5 (-9.8)(.2)^2

.544 = Vo(0.2) - .196

(.544+.196)/(.2) = Vo

Vo = 3.7 m/s

(b) Vf^2 = 3.7^2 + 2(-9.8)(.544)

Vf^2 = 13.69 - 2.67

Vf = sqrt (11.02)

Vf = 3.32 m/s

(c) 0^2 = 3.7^2+s(-9.8)(/\x)

/\x = -3.7^2/-(2*9.8)

/\x = 0.6985 m

is this correct?

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# Homework Help: Armadillo problem

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