Solve Armadillo Problem: Initial & Final Speed, Height

[Johny 5]

Homework Statement

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?

Homework Equations

/\X = Vot+.5at^2
Vf^2=Vo^2+2a/\x

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?

 

[vacuum]

Yes, your solution is correct. Great job breaking down each step and using the correct equations! Keep up the good work.