Question 96024
To find when it hits the ground, let h=0


{{{0 = -16t^2 + 70t + 8}}} 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+70*t+8=0}}} ( notice {{{a=-16}}}, {{{b=70}}}, and {{{c=8}}})


{{{t = (-70 +- sqrt( (70)^2-4*-16*8 ))/(2*-16)}}} Plug in a=-16, b=70, and c=8




{{{t = (-70 +- sqrt( 4900-4*-16*8 ))/(2*-16)}}} Square 70 to get 4900  




{{{t = (-70 +- sqrt( 4900+512 ))/(2*-16)}}} Multiply {{{-4*8*-16}}} to get {{{512}}}




{{{t = (-70 +- sqrt( 5412 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-70 +- 2*sqrt(1353))/(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 = (-70 +- 2*sqrt(1353))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-70 + 2*sqrt(1353))/-32}}} or {{{t = (-70 - 2*sqrt(1353))/-32}}}



Now break up the fraction



{{{t=-70/-32+2*sqrt(1353)/-32}}} or {{{t=-70/-32-2*sqrt(1353)/-32}}}



Simplify



{{{t=35 / 16-sqrt(1353)/16}}} or {{{t=35 / 16+sqrt(1353)/16}}}



So these expressions approximate to


{{{t=-0.11144676971869}}} or {{{t=4.48644676971869}}}



So our possible solutions are:

{{{t=-0.11144676971869}}} or {{{t=4.48644676971869}}}


However, since a negative time doesn't make sense, our only solution is 


{{{t=4.48644676971869}}}



So it will take about 4.486 seconds to hit the ground.