A bush baby is capable of leaping vertically to the remarkable height of 2.3m. T

Question

A bush baby is capable of leaping vertically to the remarkable height of 2.3m. To jump this hight, the baby acclerates over a distance of .16m while rapidly extending its legs. The accelration dureing the jump is approx. constant. Whiat is the accelrateion in m/s^2 and in g's.
Please just mention the two kinematic equations that are necessary to solve the problem and break them up. For instance, the part where the baby goes from .16-2.3 deserves one equation, and the part where you go from y=0-y=.16. This problem is only in chapter 2 in my book and therefor only has the need for the equations used for motion w/ constant accel.

Explanation / Answer

so v = v(2gh) = v(2*9.8*2.14) = 6.5m/s Then the acceleration is given as    v^2 = v0^2 + 2ax therefore the acceleration a = v^2 / 2x = (6.5m/s)^2 / 2(0.16m)    = 132 m/s^2

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