Question 120492
Your formula is incorrect.


If you drop an object from a fixed height, s, the height, h, of the object after t seconds is given by {{{h=-16t^2+s}}}.  You want the time it takes for the object starting at 1400 feet to hit the ground, or when h = 0, so:

{{{0=-16t^2+1400}}}


{{{16t^2=1400}}}


{{{t^2=1400/16}}}


{{{t=sqrt(1400/16)}}}


{{{t=sqrt(1400)/4}}}


{{{t=(5*sqrt(14))/2}}}, roughly 9 and 1/3 seconds.


In general, if you drop an object from height s in feet and you want to know the time it takes to reach height h in feet, it takes:


{{{t=sqrt(s-h)/4}}} seconds.


Also, if the heights are given in meters, then the constant factor is different:


{{{h=4.9t^2+s}}}, so


{{{t=sqrt((s-h)/4.9)}}}


Hope that helps
John