This material is encountered if you enrolled in engineering or college level
courses. High School Physics (Blackwood Herron  Kelly) has some coverage.
See College Physics by Sears and Zemansky on freely falling bodies or a
similar text and topic.

          Acceleration of gravity g = 32   ft/(sec * sec)
          Velocity v feet/sec = g * Time t  sec
          Initial condition, at time = 0, Velocity vo = 0
          Distance traveled d = vo + (1/2) * g * t * t feet
          The free fall distanced traveled if vo = 0  is
          d = (1/2) * g * t * t  feet
          The time to drop distance t = ((2 * d) / g )^(1/2) sec
          The velocity at drop of d   = ( 2 * g * d )^(1/2)  feet/sec

          Lets drop 10,000 feet and find the velocity v
          v =  ( 2 * 32 )^(.5) * (10000)^(.5)
            =  (64)^(.5) * (100 * 100)^(.5)
            =  ( 8 * 8 )^(.5) *  100
            =    800 ft/sec
         So after droping 10,000 feet we are at 800 ft/sec, if dropped
         from 10,800 feet you might say we have 1 sec before hitting the
         ground, but you would be wrong.

         How long to drop 10,000 feet is
         t =   ((2 * d) / g)^(.5)
           =   ( 4 * d / 64)^(.5)
           =     (1/8) * (4 * d)^(.5)
           =     (1/4) * (100 * 100)^(.5)
           =     (1/4) *  100
           =     25 sec
         to drop 10,800 feet
         t =     (1/4) * (10800)^(.5)
           =     (1/4) *   103.923
           =     25.98 sec
         but not far off.