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?

Answer 1

hf=hi+vi*t-4.8t^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.

Answer 2

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