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A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.26 m. To jump this high, the bush baby accelerates over a distance of 0.160 m, while extending the legs. The acceleration during the jump is approximately constant. What is the acceleration in m/s?

User Sefiks
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1 Answer

1 vote

Answer:

The acceleration from it's legs is
a=138.20(m)/(s^(2) )

Step-by-step explanation:

Let's order the information:

Initial height:
y_(i)=0m

Final height:
y_(f)=2.26m

The bush accelerates from
y_(i)=0m to
y_(e)=0.16m.

We can use the following Kinematic Equation to know the velocity at
y_(e):


v_(f)^(2) = v_(e)^(2) - 2g(y_(f)-y_(e))=2ay_(e)

where g is gravity's acceleration (9.8m/s). Since
v_(f)=0,


2g(y_(f)-y_(e)) = v_(e)^(2)


v_(e)=6.41(m)/(s)

Working with the same equation but in the first height interval:


v_(e)^(2) = v_(i)^(2) + 2(a-g)(y_(e)-y_(i))

Since
v_(i)=0 and
y_(i)=0,


v_(e)^(2) = 2(a-g)y_(e)


a-g=(v_(e)^(2))/(2y_(e))


a=(v_(e)^(2))/(2y_(e))+g
a=138.20(m)/(s^(2) )

User Rodrigo Reis
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