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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.
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