Posted by TechnoBoi11 on .
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?
(b) What is its speed at the height of 0.547 m?
(c) How much higher does it go?
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.