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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.
18 years ago

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