You can
put this solution on YOUR website!A rocket is launched from atop a 43-foot cliff with an initial velocity of 70 feet per second. The height of the rocket t seconds after launch is given by the equation h = -16t^2 + 70t + 43. Graph the equation to find out how long after the rocket is launched it ill hit the ground. Estimate your answer to the nearest tenth of a second.
------------------
h = -16t^2 + 70t + 43
Set h = 0 and solve for t
-16t^2 + 70t + 43 = 0
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=7652 is greater than zero. That means that there are two solutions: .


Quadratic expression can be factored:

Again, the answer is: -0.54611596607863, 4.92111596607863.
Here's your graph:
 |
-------------
Use the positive number.
----------------
BTW, rockets have motors and they accelerate when launched. This is a ballistics problem, and the "rocket" just rises and falls due to initial forces and gravity.
------------
For a better graph, get the FREE software from
http://www.padowan.dk.com