# physics

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

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1. 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

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2. 11.7 g

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