Question 87894
Well, you can start with the function that gives the height, as a function of time, h(t) of an object propelled upward with an initial upward velocity of {{{v[0]]]}}}m/s from an initial height of {{{h[0]}}}m.
{{{h(t) = -4.9t^2+v[0]t+h[0]}}}
In this problem:
{{{h[0] = 100}}}m.
{{{v[0] = 20}}}m/s.
You would like to know at what time, t, will h = 80 meters.
Making the appropriate substitutions into the formula:
{{{80 = -4.9t^2+20t+100}}} Subtract 80 from both sides.
{{{-4.9t^2+20t+20 = 0}}} Solve this quadratic equation using the quadratic formula: {{{t = (-b+-sqrt(b^2-4ac))/2a}}} In this equation, a = -4.9, b = 20, and c = 20
{{{t = (-20+-sqrt(20^2-4(-4.9)(20)))/2(-4.9)}}} Simplifying this, you'll get:
{{{t = (-20+-sqrt(400-(-392)))/-9.8}}}
{{{t = (-20+-sqrt(792))/-9.8}}} Evaluating this gets you:
{{{t = 4.912499}}} or {{{t = -0.830866}}} Discard the negative solution.
{{{t = 4.9}}}seconds.