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 0.18m while extending the legs. The acceleration during the jump is approximately constant.

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

11.7 g

Well, well, well, looks like the bush baby is a real high-flyer! To find the acceleration during the pushing-off phase, we can use a simple physics formula: Final Velocity squared = Initial Velocity squared + 2 * acceleration * distance.

Given that the bush baby starts from rest, we can say that the initial velocity is 0. Therefore, we have Final Velocity squared = 2 * acceleration * distance.

Plugging in the given values, we get (2.3 m/s)^2 = 2 * acceleration * 0.18 m.

Now let's do some math gymnastics. √(2.3^2 / (2 * 0.18)) = √(2.075 / 0.36) ≈ √5.764 ≈ 2.4.

So, the acceleration during the pushing-off phase is approximately 2.4 m/s². The bush baby sure knows how to launch itself into the air with some serious acceleration!

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

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, as the bush baby reaches its maximum height, its vertical velocity momentarily becomes zero)
u = initial velocity (also 0 m/s, as the bush baby starts from rest)
a = acceleration (unknown)
s = displacement (0.18 m, as given in the question)

Rearranging the equation, we have:

0 = 0^2 + 2a(0.18)

Simplifying, we get:

0 = 0 + 0.36a

Dividing both sides by 0.36, we find:

0 = a

Therefore, the acceleration during the pushing-off phase of the bush baby's jump is approximately 0 m/s^2.

Has unknown speed Vi at h = .18 m

while coasting upward:
0 = Vi - 9.8 t
so t = Vi/9.8

2.3 = .18 + Vi t - (1/2) 9.8 t^2
2.12 = Vi^2/9.8 - (1/2) Vi^2/9.8

(1/2) Vi^2 = 9.8*2.12
Vi = 6.45 m/s at .18 meters
so
average speed between ground and .18 meters = 6.45/2 = 3.22 m/s
so time to reach .18 m = .18/3.22 = .056 seconds
a = change in speed / change in time
= 6.45/.056 = 115 m/s^2

10 g, no way