Question 92897
{{{h=120+8t-16t^2}}}


{{{60=120+8t-16t^2}}} Plug in h=60


{{{0=-16t^2+8t+60}}} Subtract 60 from both sides


Let's use the quadratic formula to solve for t:



Starting with the general quadratic


{{{at^2+bt+c=0}}}


the general solution using the quadratic equation is:


{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{-16*t^2+8*t+60=0}}} ( notice {{{a=-16}}}, {{{b=8}}}, and {{{c=60}}})


{{{t = (-8 +- sqrt( (8)^2-4*-16*60 ))/(2*-16)}}} Plug in a=-16, b=8, and c=60




{{{t = (-8 +- sqrt( 64-4*-16*60 ))/(2*-16)}}} Square 8 to get 64  




{{{t = (-8 +- sqrt( 64+3840 ))/(2*-16)}}} Multiply {{{-4*60*-16}}} to get {{{3840}}}




{{{t = (-8 +- sqrt( 3904 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-8 +- 8*sqrt(61))/(2*-16)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{t = (-8 +- 8*sqrt(61))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-8 + 8*sqrt(61))/-32}}} or {{{t = (-8 - 8*sqrt(61))/-32}}}



Now break up the fraction



{{{t=-8/-32+8*sqrt(61)/-32}}} or {{{t=-8/-32-8*sqrt(61)/-32}}}



Simplify



{{{t=1 / 4-sqrt(61)/4}}} or {{{t=1 / 4+sqrt(61)/4}}}



So these expressions approximate to


{{{t=-1.70256241897666}}} or {{{t=2.20256241897666}}}



So our solutions are:

{{{t=-1.70256241897666}}} or {{{t=2.20256241897666}}}


Since a negative time doesn't make much sense, our only solution is 


{{{t=2.20256241897666}}}


which is about 2.2 seconds