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

Question

A bush baby, is capable of leaping vertically to the remarkable height of 2.3m. To jump this high, the bush baby accelerates over a distance of .16m while rapidly extending its legs. The acceleration during the jump is approx. constant. What is the acceleration in m/s^2 and the g's.

Explanation / Answer

According to conservation of energy we have Change in kinetic energy = change potential energy ==> 0.5 m v^2 - 0 = 0 + m gh ==> v^2 = 2 g h .........1 Along the horizontal direction we have v^2 = u^2 + 2 a d ==> v^2 = 0 + 2 a d .........2 From 1 and 2 we have 2 a d = 2 g h ==> a = g h / d = g * 2.3 / 0.16 = 14.37 g = 140.875 m/s^2

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