A startled armadillo leaps upward rising 0.540 m in the first 0.195 s.

Question

(a) What is its initial speed as it leaves the ground?
3.7 m/s

(b) What is its speed at the height of 0.540 m?
1.8 m/s

(c) How much higher does it go?

How would I calculate (c)? should I use the formula vf^2=vi^2+2a(yf-yi)?

Answer 1

(a) In 0.195 s, it will slow down by 9.81*0.195 = 1.913 m/s. The average speed during the interval was 2.769 m/s.
[Vo + (Vo-1.91)]/2 = 2.769
Vo - 0.956 = 2.760
Vo = 3.72 m/s
You got that right! Good show!

(b) Vo^2 - V^2 = 2 g H = 10.60 m^2/s^2
V^2 = 3.238 m^2/s^2
V = 1.80 m/s touche!

(c) It rises an additional distance H' such that
2 g H' = V^2
H' = 0.165 m

Answer 2

Thanks!