Question

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

Answer 1

Answer:

$a = 12.78 g's$

Explanation:

Height reached by the object after push off is given as

$H = 2.3 m$

$v_f^2 - v_i^2 = 2 a s$

now we have

$0 - v^2 = 2(-9.81)(2.3)$

$v = 6.72 m/s$

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

so during the push we will have

$v_f^2 - v_i^2 = 2 a d$

$6.72^2 - 0 = 2a(0.18)$

$a = 125.35 m/s^2$

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

$a = \frac{125.35}{9.81} gs$

$a = 12.78 g's$