Question 329341
Start with the function for the height (h)of an object thrown upward as a function of time (t).
{{{h(t) = -16t^2+v[0]*t+h[0]}}} Where: {{{h}}} = height, {{{t}}}= time, {{{v[0]}}} = Initial upward velocity, and {{{h[0]}}} = initial height of the object.
{{{h(t) = -16t^2+80*t+0}}} First, find the time ({{{t[m]}}}) at the maximum height.
{{{t[m] = -80/2(-16)}}} This is the time, in seconds, when the object reaches it maximum height.
{{{t[m] = 5/2}}} now substitute this into the first function {{{h(t)}}} to find the maximum height reached by the object.
{{{h(5/2) = -16(5/2)^2+80(5/2)}}} Evaluate.
{{{h(5/2) = 100}}}feet.  This is the maximum height attained by the object.
Now you need to find the time (t) at which the (h) = 0, so...
{{{0 = -16t^2+80t}}} Factor a t.
{{{t(-16t+80) = 0}}} so...
{{{-16t+80 = 0}}}
{{{-16t = -80}}} 
{{{t = 5}}}seconds. The object reaches the ground in 5 seconds.