Respuesta :
Answer:
a) 3.6 m/s
b) 1.53 m/s
c) 0.12 m
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)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.54=u\times 0.216+\frac{1}{2}\times -9.81\times 0.216^2\\\Rightarrow u=\frac{0.54+\frac{1}{2}\times 9.81\times 0.216^2}{0.216}\\\Rightarrow u=3.6\ m/s[/tex]
Initial speed as it leaves the ground is 3.6 m/s
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 0.54+3.6^2}\\\Rightarrow v=1.53\ m/s[/tex]
Speed at the height of 0.540 m is 1.53 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-3.6^2}{2\times -9.81}\\\Rightarrow s=0.66\ m[/tex]
The total height the armadillo leaps is 0.66 m
So, the additional height is 0.66-0.54 = 0.12 m
