Answer:
The acceleration from it's legs is [tex]a=138.20\frac{m}{s^{2} }[/tex]
Explanation:
Let's order the information:
Initial height: [tex]y_{i}=0m[/tex]
Final height: [tex]y_{f}=2.26m[/tex]
The bush accelerates from [tex]y_{i}=0m[/tex] to [tex]y_{e}=0.16m[/tex].
We can use the following Kinematic Equation to know the velocity at [tex]y_{e}[/tex]:
[tex]v_{f}^{2} = v_{e}^{2} - 2g(y_{f}-y_{e})=2ay_{e}[/tex]
where g is gravity's acceleration (9.8m/s). Since [tex]v_{f}=0[/tex],
[tex]2g(y_{f}-y_{e}) = v_{e}^{2}[/tex]
⇒ [tex]v_{e}=6.41\frac{m}{s}[/tex]
Working with the same equation but in the first height interval:
[tex]v_{e}^{2} = v_{i}^{2} + 2(a-g)(y_{e}-y_{i})[/tex]
Since [tex]v_{i}=0[/tex] and [tex]y_{i}=0[/tex],
[tex]v_{e}^{2} = 2(a-g)y_{e}[/tex]
⇒[tex]a-g=\frac{v_{e}^{2}}{2y_{e}}[/tex]
⇒[tex]a=\frac{v_{e}^{2}}{2y_{e}}+g[/tex] ⇒ [tex]a=138.20\frac{m}{s^{2} }[/tex]
