What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an African primate, is capable of a remarkable vertical leap. The bush baby goes into a crouch and extends its legs, pushing upward for a distance of 0.18 m. After this upward acceleration, the bush baby leaves the ground and travels upward for 2.3 m. (Figure 1) a 130 m/s All attempts used; correct answer displayed Part B Figure < 1of1 What is the acceleration during the pushing-off phase, in gs? Express your answer in terms of g Submit Provide Feedback

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aristocles aristocles · Answer 1 · 2019-10-20T07:29:33+02:00

Answer:

[tex]a = 12.78 g's[/tex]

Explanation:

Height reached by the object after push off is given as

[tex]H = 2.3 m[/tex]

[tex]v_f^2 - v_i^2 = 2 a s[/tex]

now we have

[tex]0 - v^2 = 2(-9.81)(2.3)[/tex]

[tex]v = 6.72 m/s[/tex]

now we know that this push last for total distance of 0.18 m

so during the push we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]6.72^2 - 0 = 2a(0.18)[/tex]

[tex]a = 125.35 m/s^2[/tex]

now in terms of g = 9.81 m/s/s we have

[tex]a = \frac{125.35}{9.81} gs[/tex]

[tex]a = 12.78 g's[/tex]