Question 81891
The instruction do indicate to determine the solution by graphing, so let's see what the graph looks like:
Graph the function:{{{h(t) = -16t^2+152t+78}}}
Instead of graphing x (horizontally) and y (vertically) as in the usual graph, here, we'll be graphing t (horizontally) and h (veritically).
In the resulting graph, we'll be looking for where the graph crosses the t-axis because this is when the height of the rocket will be zero (ground-level).
{{{graph(300,200,-5,15,-25,450,-16x^2+152x+78)}}}
Looking at positive t-axis only (negative time is not meaningful here) the graph indicates that the curve (parabola) crosses the t-axis at about t = 10 seconds.
The rocket will hill the ground about 10 seconds after launch.
Let's check this solution algebraically:
{{{h(t) = -16t^2+152t+78}}}
We need to find the "zeros" or the roots of this quadratic equation.
Using the quadratic formula:{{{x = (-b+-sqrt(b^2-4ac))/2a}}}
Of course, we are looking for t rather than x, so...
{{{t = (-152+-sqrt(152^2-4(-16)(78)))/2(-16)}}}
{{{t = (-152+-sqrt(28096))/-32}}}
The roots are:
{{{t = 9.988}}} and
{{{t = -0.488}}} Discard this solution as a negative time is not meaningful.