Using s = 193 t² and a 16 inch drop, we
set 16 = 193 t², then t² = 16/193 = .0829, and taking the
square root of both sides gives t = .288 sec.
So total flight time F = 2 (.288) = .576 sec.
The horizontal rate is found by: r = 16/.576 = 27.8 in/sec.
Then the horizontal model of motion is: x = 27.8 t inches.