Question 309267
The answer, of course, depends upon where the object stops.  In this problem, we'll assume that the object falls to the ground.
Using the formula for the height of a freely-falling object as a function of time:
{{{h(t) = -(1/2)Gt^2+v[0]t+h[0]}}} where: {{{h = height}}}, {{{G = 32}}}ft/sec-sq, {{{v[0] = 0}}} the initial velocity of the object, and {{{h[0] = 1815}}}ft., the initial height of the object.
When the object reaches the ground, then {{{h(t) = 0}}}, so..
{{{0 = -16t^2+0*t+1815}}}
{{{16t^2 = 1815}}}
{{{t^2 = 113.4375}}} and...
{{{highlight(t = 10.65)}}}seconds.