Question 996115
If that planet is Earth, or you meant "thrown from a plane", and the object is thrown downwards, I have an answer.
The acceleration of gravity for objects as close to Earth's surface as non-astronauts ever get is {{{g=about}}}{{{32}}}{{{ft/second^2}}} .
With an initial downward velocity of {{{16}}}{{{ft/second}}} ,
the downwards velocity (in ft/s) for a heavy enough object
would be {{{v=16+32t}}} ,
as a function of time, {{{t}}} (in seconds) since the object was thrown.


NOTE:
In real life, some objects (like a helium filled balloon) will not fall like that, because buoyancy makes the downward "weight" force less than mass times {{{g}}}.
Also, even for heavy enough objects, air drag will eventually stop the velocity increase,
and a constant velocity (terminal velocity) will be asymptotically approached.


IF WE CAN DISREGARD THE EFFECTS OF BUOYANCY AND AIR DRAG:
The average velocity for the first {{{t}}} seconds is
{{{(16+(16+32t))/2=16+16t}}} .
The distance (in ft) the object travels over {{{t}}} seconds is
{{{(16+16t)*t=16t^2+16t}}} .
When the object has traveled {{{1449ft}}} downwards,
it strikes the ground, and
{{{16t^2+16t=1449}}}<--->{{{16t^2+16t-1449=0}}} .
Solving that quadratic equation, we find
{{{t=(-16 +- sqrt(16^2-4*16*(-1449)))/(2*16)=(-16 +- sqrt(256+92736))/32=(-16 +- sqrt(92992))/32=about9.03}}}


IF WE HAVE TO CONSIDER AIR DRAG AND/OR BUOYANCY:
It will take longer for that object to strike the ground,
but it is a more complicated calculation,
and it depends on the "drag coefficient" of the object,
and that depends of the shape
With high terminal velocities of about 90 m/s (about 300 ft/s) reported (for a bullet),
it is likely that drag would be a factor, because without it,
after 9 seconds the velocity (in ft/s) would be
{{{16+32*9=314}}}ft/s.