Question 103994
The object will hit the ground when the height is 0, so let h=0 


{{{0=-16t^2+90t+10}}}



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+90*t+10=0}}} ( notice {{{a=-16}}}, {{{b=90}}}, and {{{c=10}}})





{{{t = (-90 +- sqrt( (90)^2-4*-16*10 ))/(2*-16)}}} Plug in a=-16, b=90, and c=10




{{{t = (-90 +- sqrt( 8100-4*-16*10 ))/(2*-16)}}} Square 90 to get 8100  




{{{t = (-90 +- sqrt( 8100+640 ))/(2*-16)}}} Multiply {{{-4*10*-16}}} to get {{{640}}}




{{{t = (-90 +- sqrt( 8740 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-90 +- 2*sqrt(2185))/(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 = (-90 +- 2*sqrt(2185))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-90 + 2*sqrt(2185))/-32}}} or {{{t = (-90 - 2*sqrt(2185))/-32}}}



Now break up the fraction



{{{t=-90/-32+2*sqrt(2185)/-32}}} or {{{t=-90/-32-2*sqrt(2185)/-32}}}



Simplify



{{{t=45 / 16-sqrt(2185)/16}}} or {{{t=45 / 16+sqrt(2185)/16}}}



So these expressions approximate to


{{{t=-0.108998973130061}}} or {{{t=5.73399897313006}}}



So our possible solutions are:

{{{t=-0.108998973130061}}} or {{{t=5.73399897313006}}}



However, since a negative time doesn't make sense, our only solution is {{{t=5.73399897313006}}}



So it takes about 5.7 seconds for the object to hit the ground.