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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?

Respuesta :

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

Answer: 12.75 m/s