A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.3 m. To jump this high, the bush baby accelerates over a distance of 0.14m while extending the legs. The acceleration during the jump is approximately constant.

-What is the acceleration during the pushing-off phase, in m/s2?

-What is the acceleration during the pushing-off phase, in g?

To find the acceleration during the pushing-off phase for the bush baby's jump, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

where:
vf = final velocity (0 m/s, since the bush baby reaches its maximum height and then comes back down)
vi = initial velocity (0 m/s, since the bush baby starts from rest)
a = acceleration (unknown)
d = distance (0.14 m)

Since the bush baby extends its legs during the jump, we can assume that the acceleration is constant.

Now, rearranging the equation to solve for acceleration (a):

a = (vf^2 - vi^2) / (2d)

Substituting the given values:

a = (0^2 - 0^2) / (2 * 0.14)

Simplifying:

a = 0 / 0.28

The acceleration during the pushing-off phase is 0 m/s^2.

To find the acceleration in g, we need to convert the units. 1 g is equal to 9.8 m/s^2.

So, to convert from m/s^2 to g:

Acceleration in g = acceleration in m/s^2 / 9.8

Substituting the value we found:

Acceleration in g = 0 / 9.8

The acceleration during the pushing-off phase is 0 g.