Posted by **TechnoBoi11** on Thursday, January 25, 2007 at 5:48am.

A startled armadillo leaps upward rising 0.547 m in the first 0.202 s.

(a) What is its initial speed as it leaves the ground?

_____m/s

(b) What is its speed at the height of 0.547 m?

_____m/s

(c) How much higher does it go?

_____m

Use distance vs time:

d=vi*t - 1/2 g t^2 solve for vi.

THen, for the second

d=vi*t -1/2 g t^2 solve for t, then

v=vi*t solve for v.

Finally, knowing vi, solve for the time to the top, where vf is zero.

Vf=vi*t -g t solve for t

height=vi*t - 1/2 g t^2 solve for height.

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