A startled armadillo leaps upward, rising 0.596 m in the first 0.190 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.596 m? (c) How much higher does it go?

Answer 1

hf(t)=vi*t- 1/2 g t^2
.596=vi*.190- 4.9 (.190)^2
solve for vi

b) vf=Vi-gt t=.190, Vi above

c) how high?
Vf at top is zero.
Vf^2=Vi^2+2gd solve for d.