Question 1124898

Not sure if you mean,
{{{t=sqrt(d)/2}}} or {{{t=sqrt(d/2)}}}

So,if {{{t=1.08}}} seconds, we have

{{{1.08=sqrt(d)/2}}}

{{{sqrt(d)=2(1.08)}}}...square both sides

{{{(sqrt(d))^2=(2(1.08))^2}}}


{{{d=(2.16)^2}}}

{{{d=4.7}}}  in feet