Question 119166
The ball will hit the ground when h=0



{{{h=-16t^2+80t+3}}} Start with the given equation



{{{0=-16t^2+80t+3}}}  Plug in h=0



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+80*t+3=0}}} ( notice {{{a=-16}}}, {{{b=80}}}, and {{{c=3}}})





{{{t = (-80 +- sqrt( (80)^2-4*-16*3 ))/(2*-16)}}} Plug in a=-16, b=80, and c=3




{{{t = (-80 +- sqrt( 6400-4*-16*3 ))/(2*-16)}}} Square 80 to get 6400  




{{{t = (-80 +- sqrt( 6400+192 ))/(2*-16)}}} Multiply {{{-4*3*-16}}} to get {{{192}}}




{{{t = (-80 +- sqrt( 6592 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-80 +- 8*sqrt(103))/(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(103))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-80 + 8*sqrt(103))/-32}}} or {{{t = (-80 - 8*sqrt(103))/-32}}}



Now break up the fraction



{{{t=-80/-32+8*sqrt(103)/-32}}} or {{{t=-80/-32-8*sqrt(103)/-32}}}



Simplify



{{{t=5 / 2-sqrt(103)/4}}} or {{{t=5 / 2+sqrt(103)/4}}}



So these expressions approximate to


{{{t=-0.0372228912730548}}} or {{{t=5.03722289127306}}}



So our possible solutions are:

{{{t=-0.0372228912730548}}} or {{{t=5.03722289127306}}}



However, since a negative time doesn't make sense, our only solution is {{{t=5.03722289127306}}}



So it takes about 5.03 seconds for the ball to hit the ground.