Question 68718
First, the formula for the height of an object (in meters) propelled upwards with an initial velocity of {{{v[0]}}} from an initial height of {{{h[0]}}} is given by:
{{{h(t) = -4.9t^2 + v[0]t + h[0]}}} where:
{{{v[0] = 20m/s}}}
{{{h[0] = 100m}}}
You want to find h(t)= 80, the time at which the ball reaches 80 m. You will, of course, expect to get two such times, one of which won't make any practical sense because your initial height (100m) is above the 80m point. The second time will be meaninful because the ball will be on the way down. Let's see what comes out.
{{{-4.9t^2 + 20t + 100 = 80}}} Simplify and solve for t. Subtract 80 from both sides.
{{{-4.9t^2 + 20t + 20 = 0}}} Use the quadratic formula:{{{t = (-b+-sqrt(b^2-4ac))/2a)}}}
In this case, a = -4.9, b = 20, and c = 20.
{{{t = (-20+-sqrt(20^2-4(-4.9)(20)))/2(-4.9)}}} Simplify.
{{{t = (-20+-sqrt(400-(-392)))/-9.8)}}}
{{{t = (-20+-sqrt(792))/-9.8}}}
{{{t = 4.9}}}seconds. This is the solution.
{{{t = -0.83}}} Discard this as the time must be positive.