Question 42657
{{{h = -16t^2 + 48t}}}


When the ball will return to ground,  h = 0; you are right.
Put h = 0 in the above equation (You have commited some error in this step.)


{{{0 = -16t^2 + 48t}}}
or {{{16t^2 - 48t = 0}}} [This is the quadratic equation; {{{red(a = 16)}}}, {{{red(b = -48)}}}, {{{red(c = 0)}}}]
or {{{16t(t-3) = 0}}}
or {{{t(t-3) = 0/16}}}
or {{{t(t-3) = 0}}}


As product of two quantities = 0, so at least one of them must be zero.
Hence either t = 0 or (t - 3) = 0 i.e. either t = 0 or t = 3.
Now, at t=0 the ball was thrown up so then its height was also zero.
So t = 0 cannot be the time when it comes down. 
So t = 3 must be the time when it comes down.


The ball was thrown up at time t = 0, it came down at time t = 3.
So time of flight = (time when it comes down) - (time when it was thrown) = 3 - 0 = 3.


So the answer is 3 seconds.
Got it now?