Question 88620
{{{f(x) = (-1/2)gx^2 + vx + h}}}
f(x) determines height after x seconds
g is the gravitational constant
v is the initial velocity
h is the initial height
{{{80 = (-1/2)32x^2 + 20x + 100}}}
{{{-20 = -16x^2 + 20x}}}
{{{5/4 = x^2 - 5x/4)}}}
{{{105/64 = 80/64 + 25/64 = 5/4 + 25/64 = (x - 5/8)^2}}}
{{{0 +- sqrt(105)/8 = x - 5/8}}}
{{{(5 +- sqrt(105))/8 = x}}}
After {{{(5 + sqrt(105))/8}}} seconds, the ball would be 80m high.
{{{graph(300,400,-3,4,-20,120,80,-16x^2 + 20x + 100)}}}