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 0.14m while extending the legs. The acceleration during the jump is approximately constant.
What is the acceleration during the pushing-off phase, in m/s2?
Please i need help ASAP :(
To find the acceleration during the pushing-off phase, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s when the bush baby reaches the maximum height)
u = initial velocity (which is 0 m/s since the bush baby starts from rest)
a = acceleration
s = distance (0.14 m)
We rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Since v=0 and u=0, the equation simplifies to:
a = 0 / (2s) = 0
Therefore, the acceleration during the pushing-off phase is 0 m/s^2.
It's important to note that the bush baby accelerates to its maximum velocity during the pushing-off phase but doesn't continue accelerating once it leaves the ground, resulting in an acceleration of zero during this phase.