Aball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resistance can be ignored, how high does it go? \

Convert the given in SI units.

         (44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec

The distance traveled and the initial velocity can be related through the equation,
  
   d = (Vf)² - (Vi)²/ 2a

where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,

    d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)

The value of d from the equation,

    d = 9.17 meters

Convert this to feet,

    d = (9.17 m)(3.28 ft / 1 m) = 30 ft

30 ft