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 11.7 ft when leaving the ground at an angle of 41.1°. with what speed, in si units, does it leave the ground to make this leap?

The launch velocity is at 41.1° relative to the horizontal.
Let V = the launch velocity, ft/s
The horizontal component of the velocity is
Vx =Vcos(41.1°) = 0.7536V ft/s
The vertical component of the launch velocity is
Vy = Vsin(41.1) = 0.6574V ft/s

Assume that g = 32 ft/s² and neglect air resistance. 

At the maximum height of 11.7 ft, the vertical velocity is zero. Therefore
Vy² - 2*32*11.7 = 0
Vy² = 748.8
Vy = 27.3642 ft/s

Therefore
0.6574V = 27.3642
V = 41.8285 ft/s

Note that
1 ft/s = 0.3048 m/s
Therefore
V = 41.8285*0.3048 = 12.749 m/s

12.75 m/s

If there was no gravity then their would be no downword force on the football to have it return the the grownd, and if there was no friction (mainly air friction) then there would be no opposing force the stop the ball from moving through the air allowing the ball to travle ∞.