SOLUTION: Hello, I wanted to see if I was doing this problem correctly. I started the answer with -16t^2+16t+410=90, then subtracted 90 to get -16t^2+16t+320=0 and then perform the qua

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Hello, I wanted to see if I was doing this problem correctly. I started the answer with -16t^2+16t+410=90, then subtracted 90 to get -16t^2+16t+320=0 and then perform the qua      Log On


   



Question 682133: Hello,
I wanted to see if I was doing this problem correctly.
I started the answer with -16t^2+16t+410=90, then subtracted 90 to get -16t^2+16t+320=0 and then perform the quadratic equation to get the answer. Is this the right start?

The function h(t) = -16^2+16t+410 gives the height h, in feet, of a coin tossed upward from a balcony 410 ft high with an initial velocity of 16 ft/sec. During what interval of time will the coin be at a height of at least 90 ft?
Thanks

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
I started the answer with -16t^2+16t+410=90, then subtracted 90 to get -16t^2+16t+320=0 and then perform the quadratic equation to get the answer. Is this the right start?
Yes, applying the "quadratic equation" is one approach.
But, you could have "factored" it (an easier approach):
-16t^2+16t+320=0
dividing both sides by -16:
t^2-t-20=0
factoring:
(t+4)(t-5) = 0
t = {-4, 5}
throw away the negative solution (extraneous) leaving:
t = 5 seconds
.