Posted by Ethan on Sunday, January 16, 2011 at 1:54pm.
A startled armadillo leaps upward, rising 0.515 m in the first 0.203 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.515 m? (c) How much higher does it go?

Physics  bobpursley, Sunday, January 16, 2011 at 1:57pm
hf=hi+vi*t4.8t^2 you know hi=0, hf, t solve for vi
at the top, vf=0
vf^2=vi^2+2gh where g=9.8 solve for h.

Physics  Damon, Sunday, January 16, 2011 at 2:01pm
h = Vi t  4.9 t^2
.515 = Vi (.203) 4.9 (.041)
.203 Vi = .717
Vi = 3.53 m/s
(b) v = Vi  9.8 t
(c) 0 = Vi  9.8 t solve for t at top
then
h = Vi t  4.9 t^2
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