Question

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 13.7 ft when leaving the ground at an angle of 42.7°. with what speed, in si units, does it leave the ground to make this leap? m/s

Answer 1

Let v = the speed with which the animal leaves the ground.

Because the angle is 42.7°. the vertical launch velocity is
v sin(42.7°) = 0.6782v ft/s

Ignore air resistance, and g = 32.2 ft/s².

At maximum height, the vertical velocity is zero.
Because the maximum height is 13.7 ft, therefore
(0.6782v ft/s)² - 2*(32.2 ft/s²)*(13.7 ft) = 0
0.46v² = 882.28
v = 43.795 ft/s

Note that
1 ft/s = 0.3048 m/s
Therefore
v = 43.795*0.3048 = 13.349 m/s

Answer: 13.5 m/s (nearest tenth)