Question 98246
To find when it hits the ground, let h=0


{{{0=-16t^2 + 80t + 5}}}



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 {{{0=-16*t^2+80*t+5}}} ( notice {{{a=-16}}}, {{{b=80}}}, and {{{c=5}}})


{{{t = (-80 +- sqrt( (80)^2-4*-16*5 ))/(2*-16)}}} Plug in a=-16, b=80, and c=5




{{{t = (-80 +- sqrt( 6400-4*-16*5 ))/(2*-16)}}} Square 80 to get 6400  




{{{t = (-80 +- sqrt( 6400+320 ))/(2*-16)}}} Multiply {{{-4*5*-16}}} to get {{{320}}}




{{{t = (-80 +- sqrt( 6720 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-80 +- 8*sqrt(105))/(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 = (-80 +- 8*sqrt(105))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-80 + 8*sqrt(105))/-32}}} or {{{t = (-80 - 8*sqrt(105))/-32}}}



Now break up the fraction



{{{t=-80/-32+8*sqrt(105)/-32}}} or {{{t=-80/-32-8*sqrt(105)/-32}}}



Simplify



{{{t=5 / 2-sqrt(105)/4}}} or {{{t=5 / 2+sqrt(105)/4}}}



So these expressions approximate to


{{{t=-0.0617376914898995}}} or {{{t=5.0617376914899}}}



So our possible solutions are:

{{{t=-0.0617376914898995}}} or {{{t=5.0617376914899}}}




However, since a negative time doesn't make sense, our only solution is {{{t=5.0617376914899}}}