SOLUTION: The equation is h(t)= -16t^2+50t+4. (h for feet and t for seconds). The question is When will the ball be 40 feet above the ground. I plugged 40 into h(t), subtracted it and insert

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The equation is h(t)= -16t^2+50t+4. (h for feet and t for seconds). The question is When will the ball be 40 feet above the ground. I plugged 40 into h(t), subtracted it and insert      Log On


   

Question 838361: The equation is h(t)= -16t^2+50t+4. (h for feet and t for seconds). The question is When will the ball be 40 feet above the ground. I plugged 40 into h(t), subtracted it and inserted it into the other side. I factored but got t=9/8 and t=2. I do not think I am supposed to have two positive numbers.
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
?
?

-16t%5E2%2B50t%2B4=40
-16t%5E2%2B50t%2B4-40=0
-16t%5E2%2B50t-36=0
8t%5E2-25t%2B18=0
-
discriminant is highlight%2825%5E2-4%2A8%2A18=49%29
t=%2825%2B-+7%29%2F%282%2A8%29, TWO POSITIVE VALUES.
-
Continuing, t=%2825%2B-+7%29%2F16
t=9%2F8 or t=2
-
What goes up also comes down.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is h(t)= -16t^2+50t+4. (h for feet and t for seconds). The question is When will the ball be 40 feet above the ground. I plugged 40 into h(t), subtracted it and inserted it into the other side. I factored but got t=9/8 and t=2. I do not think I am supposed to have two positive numbers.
-----
Yes you are.
The ball starts at a height of 4 ft.
It rises at a rate of 50 ft/sec
It takes 9/8 = 1 1/8 seconds to reach a height of 40 ft because
gravity is pulling it down at 16 ft/(sec^2).
It continues to rise, then descends to a height of 40 ft.
after being in flight 2 seconds.
Cheers,
Stan H.
====================