Mayan kings and many school sports teams are named for the puma, cougar, or mountain lion felis concolor, the best jumper among animals. it can jump to a height of 13.8 ft when leaving the ground at an angle of 42.2°. with what speed, in si units, does it leave the ground to make this leap?

To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2°)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.

Iagree c would be the answer