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.

 

[thomate1]

Your solution appears to be correct. However, it would be helpful to include units in your calculations and final answers. Additionally, you may want to include a brief explanation of the equations you used and how they relate to the problem. Overall, good job on solving the armadillo problem!