SOLUTION: Alice throws a ball to Connor. The height, h (in meters) after time, t (in seconds) is approximated by the formula h= 2 + 28t – 5t^2. Calculate the time required for the ball to

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Alice throws a ball to Connor. The height, h (in meters) after time, t (in seconds) is approximated by the formula h= 2 + 28t – 5t^2. Calculate the time required for the ball to       Log On


   

Question 211092: Alice throws a ball to Connor. The height, h (in meters) after time, t (in
seconds) is approximated by the formula h= 2 + 28t – 5t^2. Calculate the time
required for the ball to reach a height of 15m.
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
First you plug in the 15 into the equation which gives you 15 = 2 + 28t - 5t^2. Now all we have to do is solve for t. To start I suggest we get everything on the same side of the equation.

15 = 2 + 28t - 5t^2
15-2-28t+5t^2 = 2-2 + 28t-28t-5t^2+5t^2
13-28t+5t^2 = 0

Now lets just rearrange the equation so its in standard form.
5t^2-28t+13 = 0

Now try as we might this equation won't factor evenly so we have to use the quadratic equation.

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ where a = 5, b = -28, and c = 13. Plugging in those values gives us:

t+=+%2828+%2B-+sqrt%28+%28-28%29%5E2-4%2A5%2A13+%29%29%2F%282%2A5%29+

Now we just simplify:

t+=+%2828+%2B-+sqrt%28+%28-28%29%5E2-4%2A5%2A13+%29%29%2F%282%2A5%29+
t+=+%2828+%2B-+sqrt%28+784-260+%29%29%2F%2810%29+
t+=+%2828+%2B-+sqrt%28+524+%29%29%2F%2810%29+
t+=+%2828+%2B-+2%2Asqrt%28+131+%29%29%2F%2810%29+
t+=+%2814+%2B-+sqrt%28+131+%29%29%2F%285%29+

Plugging this into our calculator we get answers of
t= 5.089104628 seconds or t = .510895 and I'm not sure which one we should use.