Question 52249
Substitute 0 in for h and solve:  {{{0=(-16t^2)+560t}}}.  Now factor that into {{{0=-16t(t-35)}}}.  That means that either {{{-16t=0}}} or {{{t-35=0}}}, as either factor can be 0.  Solving for each shows us that for the first equation that {{{t=0}}}, and for the second {{{t=35}}}.