Answer:
a) 3.6 m/s
b) 1.53 m/s
c) 0.12 m
Step-by-step explanation:
t = Time taken = 0.216 s
u = Initial velocity
v = Final velocity
s = Displacement = 0.54 m
a = Acceleration due to gravity = 9.81 m/s² (negative upward)

Initial speed as it leaves the ground is 3.6 m/s

Speed at the height of 0.540 m is 1.53 m/s

The total height the armadillo leaps is 0.66 m
So, the additional height is 0.66-0.54 = 0.12 m