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 0.16 m while extending the legs. The acceleration during the jump is approximately constant. What is the acceleration in m/s squared?
140 m/s^2
well, the energy the body delivers to the gravity is mgh. That has to equal force*distance= ma*distance
mgh=ma*distance
g*2.3=a*.16
solve for a.
To find the acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v is the final velocity (which is zero at the peak of the jump),
u is the initial velocity (which is the velocity at the start of the jump, zero in this case since the bush baby starts from rest),
a is the acceleration, and
s is the distance covered.
In this case, the final velocity is zero because the bush baby momentarily stops when it reaches the peak of its jump. The initial velocity is zero because the bush baby starts from rest. The distance covered during the jump is given as 0.16 m.
Plugging these values into the equation, we have:
0^2 = 0^2 + 2a(0.16)
Simplifying the equation, we get:
0 = 0 + 0.32a
Since the term "0 + 0.32a" is equal to zero, it means that the acceleration is zero. This implies that the bush baby does not experience any constant acceleration during the jump.