Question

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?

Answer 1

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) )`