A ball is projected in to the air. Its height at time t is given by the equation h = -16t^2 + 60t + 1. Solve for the time as which the height will be 8 feet. Show your work and explain the steps you used to solve. Round your answer to the nearest hundredth.

The equation gives the height of the ball. That is, h is the height of the ball. t is the time. Since we are looking for the time at which the height is 8 (h=8), we need to set the equation equal to 8 and solve for t. We do this as follows:

[tex]h=-16 t^{2} +60t+1[/tex]
[tex]8=-16 t^{2} +60t+1[/tex]
[tex]16 t^{2} -60t-1+8=0[/tex]
[tex]16 t^{2} -60t+7=0[/tex]

This is a quadratic equation and as it is set equal to 0 we can solve it using the quadratic formula. That formula is:
[tex]t= \frac{-bplus minus \sqrt{ b^{2}-4ac } }{2a} [/tex]
You might recall seeing this as "x=..." but since our equation is in terms of t we use "t-=..."

In order to use the formula we need to identify a, b and c.
a = the coefficient (number in front of) [tex] t^{2} [/tex] = 16.
b = the coefficient of t = -60
c = the constant (the number that is by itself) = 7

Substituting these into the quadratic formula gives us:
[tex]t= \frac{-(-60)plus minus \sqrt{ (-60)^{2}-4(16)(7) } }{2(16)} [/tex]
[tex]t= \frac{60plus minus \sqrt{3600-448 } }{32} [/tex]
[tex]t= \frac{60plus minus \sqrt{3152 } }{32}[/tex]

As we have "plus minus" (this is usually written in symbols with a plus sign over a minus sign) we split the equation in two and obtain:
[tex]t= \frac{60+56.1426}{32} =3.63[/tex]
and
[tex]t= \frac{60-56.1426}{32} =.12[/tex]

So the height is 8 feet at t = 3.63 and t=.12

It should make sense that there are two times. The ball goes up, reaches it's highest height and then comes back down. As such the height will be 8 at some point on the way up and also at some point on the way down.