Question 79829
The height (as a function of time) of an object propelled upward from an initial height of {{{h[0]}}} with an initial velocity of {{{v[0]}}} is given by:
{{{h(t) = -4.9t^2+v[0]t+h[0]}}}
In your problem:
{{{v[0] = 20}}}m/s
{{{h[0] = 100}}}m
You want to find at what value of t (time) will h (height) be 80 m. So, in the formula, you would set h(t) = 80 and solve for t.
{{{80 = -4.9t^2+20t+100}}} Subtract 80 from both sides.
{{{-4.9t^2+20t+20 = 0}}}  Solve for t using the quadratic formula:{{{t = (-b+-sqrt(b^2-4ac))/2a}}} where:
{{{a = -4.9}}}
{{{b = 20}}}
{{{c = 20}}} Making the appropriate substitutions, you'll get:
{{{t = (-20+-sqrt((20)^2-4(-4.9)(20)))/2(-4.9)}}}
{{{t = (-20+-sqrt(400+392))/-9.8}}}
{{{t = (-20+-sqrt(792))/(-9.8)}}}
{{{t = (-20/(-9.8))+28.14/(-9.8)}}} or {{{t = (-20/(-9.8))-28.14/(-9.8)}}}
{{{t = -0.831}}} or {{{t = 4.913}}} Discard the negative solution and keep the positive one.
The ball will reach a height of 80 meters about 4.9 seconds.