We'll ignore the air resistance. Also I suppose that "dropped" means that the initial speed is zero.
Then the speed `V` of a ball increases uniformly in magnitude under the constant gravity force, `V=-g*t` (with respect to an upward axis).
The height `H(t)` is `H_0-g*(t^2)/2` where `H_0` is the initial height and `g=32 ft/s^2` is the gravity acceleration (I think you know this). `t` is for time in seconds.
A ball reaches the ground when its height is zero, so we have to solve the equation `H(t)=0` for `tgt0.` It is simple:
`H_0-g*(t^2)/2=0,` `H_0=g*(t^2)/2,` so `t=sqrt(2H_0/g) =5.5 (s).` This is the answer.