SOLUTION: A model rocket is launched upward with an initial velocity of 240 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 240t.

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Question 1144427: A model rocket is launched upward with an initial velocity of 240 feet per second. The height, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 240t.
How many seconds after the launch will the rocket be 300 feet above the ground? Round to the nearest tenth of a second.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
To solve the problem, you need to solve the equation


    h = 300,


which is


    -16t^2 + 240t = 300,    or


    16t^2 -240t + 300 = 0


     4t^2 - 60t + 75 = 0.


Solve it using quadratic formula.

The discriminant   b^2 - 4ac = (-60)^2 - 4*4*75 = 2400 is positive, so you will have two different solutions.


One of them is the time when the body moves up; the other is the time when the body falls down.

Happy calculations (!)


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